I85PREP
⌘ K
UPMSP Intermediate Chemistry study materials
UPMSP Intermediate Chemistry study resources

UPMSP Intermediate Chemistry practice questions

One question at a time. Build your UPMSP Intermediate Chemistry understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Solutions

According to Raoult's law, the partial vapour pressure of a volatile component in an ideal solution is equal to the vapour pressure of the pure component multiplied by which quantity?

  1. Its molarity in the solution
  2. Its molality in the solution
  3. Its mass percentage in the solution
  4. Its mole fraction in the solution Correct answer.

Explanation

Raoult's law states that for an ideal solution the partial vapour pressure of each volatile component equals the vapour pressure of the pure component times its mole fraction in the liquid mixture. The law is expressed in terms of mole fraction because vapour pressure depends on the relative number of molecules, not on mass or volume-based concentration units.

Question 2Solutions

Which concentration term is defined as the number of moles of solute dissolved in one kilogram of solvent?

  1. Molality Correct answer.
  2. Molarity
  3. Normality
  4. Mole fraction

Explanation

Molality is the number of moles of solute per kilogram of solvent, and because it uses mass of solvent it does not change with temperature. Molarity, by contrast, uses litres of solution and therefore varies as the solution expands or contracts with temperature.

Question 3Solutions

A binary liquid mixture shows a large positive deviation from Raoult's law. Which statement correctly describes such a system?

  1. The total vapour pressure is lower than that predicted by Raoult's law and the mixture shows a negative enthalpy of mixing
  2. The components have identical intermolecular forces so the solution behaves ideally at all compositions
  3. The total vapour pressure is higher than predicted by Raoult's law and the enthalpy of mixing is positive Correct answer.
  4. The mixture forms a maximum-boiling azeotrope at a composition where solute-solvent interactions are strongest

Explanation

A positive deviation means the molecules escape into the vapour phase more readily than Raoult's law predicts, so total vapour pressure is higher than the ideal value. This happens when A-B attractions are weaker than the A-A and B-B attractions, and breaking the stronger pure-liquid interactions makes the mixing process endothermic with a positive enthalpy of mixing.

Question 4Solutions

Which colligative property is directly used in the laboratory to determine the molar mass of a macromolecule such as a polymer?

  1. Relative lowering of vapour pressure
  2. Osmotic pressure Correct answer.
  3. Elevation of boiling point
  4. Depression of freezing point

Explanation

Osmotic pressure is preferred for macromolecules because even a very dilute solution gives a measurable pressure, whereas boiling-point elevation and freezing-point depression become too small to read accurately at low concentration. Since osmotic pressure depends on the number of particles, it gives reliable molar masses for large molecules.

Question 5Solutions

The Henry's law constant of a gas in water is 1.0 × 10^5 kPa at a given temperature. If the partial pressure of the gas above the solution is doubled, the solubility of the gas in the liquid at equilibrium will:

  1. Become half of its original value
  2. Become four times its original value
  3. Become twice its original value Correct answer.
  4. Remain exactly the same

Explanation

Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure above the solution, with the Henry's law constant as the proportionality factor. Therefore doubling the partial pressure doubles the mole fraction of dissolved gas at equilibrium, provided temperature stays constant.

Question 6Solutions

Which pair of liquids is most likely to form an ideal solution when mixed in all proportions?

  1. Chloroform and acetone
  2. Water and nitric acid
  3. Benzene and toluene Correct answer.
  4. Water and ethanol

Explanation

Benzene and toluene are structurally similar non-polar hydrocarbons with comparable intermolecular forces, so their A-B interactions resemble the A-A and B-B interactions and the mixture closely follows Raoult's law. Water with ethanol, chloroform with acetone, and water with nitric acid all show significant deviations caused by hydrogen bonding or dipole interactions.

Question 7Solutions

A solution is prepared by dissolving 0.10 mol of a non-volatile, non-electrolyte solute in 0.90 mol of a volatile solvent. If the vapour pressure of the pure solvent is 40 mm Hg, what is the vapour pressure of the solution?

  1. 4.0 mm Hg
  2. 40 mm Hg
  3. 36 mm Hg Correct answer.
  4. 44 mm Hg

Explanation

The mole fraction of solvent is 0.90 divided by 1.00, which equals 0.90. By Raoult's law the solution vapour pressure is 0.90 times 40 mm Hg, giving 36 mm Hg, a lowering of 4 mm Hg caused by the non-volatile solute.

Question 8Solutions

Which statement about an azeotropic mixture is correct?

  1. It can be separated completely by simple distillation because the components have different boiling points
  2. It boils at a constant temperature and distils without change in composition Correct answer.
  3. Its composition in the vapour phase differs from that in the liquid phase at the boiling point
  4. It consists of two immiscible liquids that separate into layers on standing

Explanation

At an azeotropic composition the vapour has the same composition as the liquid, so boiling occurs at a constant temperature and the distillate has the same composition as the boiling mixture. Because of this, an azeotrope cannot be separated into its pure components by ordinary fractional distillation.

Question 9Solutions

The van't Hoff factor for a dilute aqueous solution of potassium sulphate, assuming complete dissociation, is:

  1. 3 Correct answer.
  2. 2
  3. 4
  4. 1

Explanation

Potassium sulphate dissociates into two potassium ions and one sulphate ion, giving three particles per formula unit. The van't Hoff factor is therefore 3, which is why its colligative effect is three times that of an equal molality of a non-electrolyte.

Question 10Solutions

Which statement correctly explains why the freezing point of a solvent is lowered when a non-volatile solute is dissolved in it?

  1. The solute reacts chemically with the solvent and changes its freezing point permanently
  2. The solute raises the vapour pressure of the solid solvent, shifting the equilibrium to a lower temperature
  3. The solute increases the rate at which the solvent crystallises out of solution
  4. The solute lowers the vapour pressure of the liquid, so the liquid and solid vapour-pressure curves intersect at a lower temperature Correct answer.

Explanation

Dissolving a non-volatile solute lowers the vapour pressure of the liquid phase while the vapour pressure of the solid solvent is unaffected. The temperature at which the two vapour-pressure curves meet, which is the freezing point, therefore shifts to a lower value, giving freezing-point depression.

Question 11Solutions

Two solutions are placed in a closed bell jar: beaker A contains pure water and beaker B contains a dilute aqueous glucose solution. After some time, what is observed?

  1. The volumes in both beakers remain unchanged because the jar is closed
  2. The volume of water in beaker A decreases and that in beaker B increases Correct answer.
  3. The volume of water in beaker B decreases and that in beaker A increases
  4. Both beakers lose water at equal rates until they become dry

Explanation

Pure water has a higher vapour pressure than the glucose solution, so water evaporates from beaker A faster than it condenses into it, and the vapour condenses preferentially into the solution in beaker B. The net result is transfer of water from the pure solvent into the solution, which is the same principle that drives osmosis.

Question 12Solutions

The osmotic pressure of a solution is measured at two different temperatures while the composition is kept the same. Which relationship correctly describes the dependence of osmotic pressure on temperature?

  1. Osmotic pressure is independent of temperature for a fixed composition
  2. Osmotic pressure decreases as temperature rises because solubility falls
  3. Osmotic pressure is inversely proportional to absolute temperature
  4. Osmotic pressure is directly proportional to absolute temperature Correct answer.

Explanation

The van't Hoff equation for osmotic pressure is written as pressure equals concentration times the gas constant times absolute temperature, so at fixed concentration the pressure rises in direct proportion to absolute temperature. This parallels the ideal gas law, reflecting the kinetic origin of osmotic pressure.

Question 13Solutions

A 0.5 molal aqueous solution of a solute shows a freezing-point depression that is 1.5 times the value expected for a non-electrolyte of the same molality. What is the most reasonable interpretation?

  1. The solute dissociates partially, giving an average van't Hoff factor of 1.5 Correct answer.
  2. The solute associates in solution, reducing the number of particles
  3. The solute is completely undissociated and the excess depression is experimental error
  4. The solvent has an unusually low cryoscopic constant

Explanation

Freezing-point depression is proportional to the total number of dissolved particles, so a depression 1.5 times the expected value means the solute produces on average 1.5 particles per formula unit. This corresponds to partial dissociation with an average van't Hoff factor of 1.5, as seen for weak electrolytes.

Question 14Solutions

Which factor is primarily responsible for the solubility of most ionic solids in water?

  1. The low dielectric constant of water, which weakens ionic bonds
  2. The non-polar nature of water, which dissolves ions without interaction
  3. Hydration of the ions, which releases energy and stabilises the separated ions Correct answer.
  4. The high density of water, which physically separates the ions

Explanation

Water has a high dielectric constant and forms strong ion-dipole interactions, so ions become surrounded by a hydration shell that releases energy and stabilises them in solution. This hydration energy helps compensate for the lattice energy that must be overcome when the ionic solid dissolves.

Question 15Electrochemistry

विद्युत-अपघटनी सेल में विद्युत धारा के प्रवाह की दिशा तथा इलेक्ट्रॉनों के प्रवाह की दिशा के बारे में कौन-सा कथन सही है?

  1. बाह्य परिपथ में इलेक्ट्रॉन ऐनोड से कैथोड की ओर जाते हैं और सेल के भीतर धारा भी ऐनोड से कैथोड की ओर मानी जाती है
  2. बाह्य परिपथ में इलेक्ट्रॉन कैथोड से ऐनोड की ओर जाते हैं और सेल के भीतर धारा ऐनोड से कैथोड की ओर मानी जाती है
  3. बाह्य परिपथ में इलेक्ट्रॉन ऐनोड से कैथोड की ओर जाते हैं और सेल के भीतर धारा कैथोड से ऐनोड की ओर मानी जाती है Correct answer.
  4. बाह्य परिपथ में इलेक्ट्रॉन कैथोड से ऐनोड की ओर जाते हैं और सेल के भीतर धारा भी कैथोड से ऐनोड की ओर मानी जाती है

Explanation

विद्युत-अपघटनी सेल में ऐनोड पर ऑक्सीकरण होता है, अतः इलेक्ट्रॉन बाह्य परिपथ से ऐनोड से निकलकर कैथोड तक पहुँचते हैं। परिपाटी के अनुसार धारा की दिशा इलेक्ट्रॉन-प्रवाह के विपरीत होती है, इसलिए बाह्य परिपथ में धारा कैथोड से ऐनोड की ओर तथा सेल के भीतर आयनों के गमन से धारा ऐनोड से कैथोड की ओर मानी जाती है।

Question 16Electrochemistry

किसी धातु M के लिए अर्ध-अभिक्रिया M²⁺ + 2e⁻ → M का मानक अपचयन विभव −0.76 V दिया गया है। इस अर्ध-सेल का मानक ऑक्सीकरण विभव कितना होगा?

  1. +0.76 V Correct answer.
  2. +1.52 V
  3. −1.52 V
  4. −0.76 V

Explanation

किसी अर्ध-अभिक्रिया का ऑक्सीकरण विभव उसी अभिक्रिया के अपचयन विभव का ऋणात्मक मान होता है, क्योंकि अभिक्रिया की दिशा उलटने पर चिह्न भी उलट जाता है। अतः E°(ऑक्सीकरण) = −E°(अपचयन) = −(−0.76 V) = +0.76 V।

Question 17Electrochemistry

सेल अभिक्रिया Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) के लिए E°(सेल) का मान क्या है, जब E°(Zn²⁺/Zn) = −0.76 V तथा E°(Cu²⁺/Cu) = +0.34 V हैं?

  1. +0.42 V
  2. +1.10 V Correct answer.
  3. −0.42 V
  4. −1.10 V

Explanation

E°(सेल) = E°(कैथोड) − E°(ऐनोड), जहाँ कैथोड पर अपचयन तथा ऐनोड पर ऑक्सीकरण होता है। यहाँ Cu²⁺ का अपचयन कैथोड पर और Zn का ऑक्सीकरण ऐनोड पर होता है, अतः E°(सेल) = 0.34 − (−0.76) = +1.10 V।

Question 18Electrochemistry

किसी गैल्वेनिक सेल में संतुलन पर नर्न्स्ट समीकरण के अनुसार कौन-सी स्थिति प्राप्त होती है?

  1. E(सेल) = E°(सेल) तथा Q = 1
  2. E(सेल) = E°(सेल) तथा Q = K
  3. E(सेल) = 0 तथा Q = 1
  4. E(सेल) = 0 तथा अभिक्रिया-भागफल Q = K Correct answer.

Explanation

संतुलन पर सेल से कोई नेट विभव उपलब्ध नहीं रहता, अतः E(सेल) शून्य हो जाता है और अभिक्रिया-भागफल Q साम्य स्थिरांक K के बराबर हो जाता है। नर्न्स्ट समीकरण में E = 0 तथा Q = K रखने पर E° = (RT/nF) ln K संबंध प्राप्त होता है।

Question 19Electrochemistry

विद्युत-अपघटनी सेल में जलीय NaCl विलयन के विद्युत-अपघटन के दौरान कैथोड पर मुख्यतः कौन-सा उत्पाद बनता है?

  1. हाइड्रोजन गैस Correct answer.
  2. ऑक्सीजन गैस
  3. सोडियम धातु
  4. क्लोरीन गैस

Explanation

जलीय विलयन में Na⁺ की तुलना में जल के H⁺ आयन अधिक सरलता से अपचयित होते हैं, क्योंकि जल का अपचयन विभव सोडियम के अपचयन विभव से अधिक धनात्मक है। अतः कैथोड पर जल का अपचयन होकर हाइड्रोजन गैस तथा हाइड्रॉक्साइड आयन बनते हैं।

Question 20Electrochemistry

25 °C पर नर्न्स्ट समीकरण में प्रयुक्त गुणक 2.303RT/F का निकटतम मान क्या है?

  1. 8.314 V
  2. 0.118 V
  3. 0.0591 V Correct answer.
  4. 0.0257 V

Explanation

25 °C (298 K) पर 2.303RT/F का मान लगभग 0.0591 V होता है, जिससे नर्न्स्ट समीकरण E = E° − (0.0591/n) log Q के रूप में लिखा जाता है। यह गुणक ताप पर निर्भर करता है, इसलिए अन्य तापों पर इसका मान बदल जाता है।

Question 21Electrochemistry

किसी सेल अभिक्रिया का मानक सेल विभव E°(सेल) = +0.46 V है तथा इसमें 2 इलेक्ट्रॉनों का स्थानांतरण होता है। 25 °C पर साम्य स्थिरांक log K का मान लगभग कितना होगा?

  1. 7.8
  2. 3.9
  3. 0.46
  4. 15.6 Correct answer.

Explanation

25 °C पर E°(सेल) = (0.0591/n) log K, अतः log K = nE°/0.0591 = (2 × 0.46)/0.0591 ≈ 15.6। इससे ज्ञात होता है कि अभिक्रिया का साम्य स्थिरांक अत्यधिक बड़ा है, अर्थात अभिक्रिया लगभग पूर्णतः उत्पादों की ओर जाती है।

Question 22Electrochemistry

हाइड्रोजन इलेक्ट्रोड के विभव पर pH का प्रभाव किस प्रकार व्यक्त किया जाता है, जब H⁺ का दाब 1 bar है?

  1. E = +0.0591/pH
  2. E = −0.0591 × pH Correct answer.
  3. E = 0.0591 × pH
  4. E का pH से कोई संबंध नहीं

Explanation

हाइड्रोजन इलेक्ट्रोड के लिए E = E° − (0.0591/1) log(1/[H⁺]) लिखा जाता है, जहाँ E° = 0। चूँकि log(1/[H⁺]) = pH, अतः E = −0.0591 × pH प्राप्त होता है, इसलिए pH बढ़ने पर विभव ऋणात्मक दिशा में बदलता है।

Question 23Electrochemistry

शुष्क सेल (लेक्लांशे सेल) के संदर्भ में कौन-सा कथन सही है?

  1. इसमें जिंक ऐनोड होता है, कार्बन छड़ कैथोड का कार्य करती है और यह प्राथमिक सेल है Correct answer.
  2. इसमें दोनों इलेक्ट्रोड जिंक के बने होते हैं तथा यह द्वितीयक सेल है
  3. इसमें जिंक ऐनोड होता है तथा विद्युत-अपघट्य केवल शुद्ध जल होता है
  4. इसमें जिंक कैथोड तथा कार्बन ऐनोड होता है और यह पुनः आवेशित की जा सकती है

Explanation

शुष्क सेल में जिंक का बेलन ऐनोड होता है जहाँ ऑक्सीकरण होता है, और केंद्र में स्थित कार्बन छड़ कैथोड का कार्य करती है। इसमें अभिक्रियाएँ अपरिवर्तनीय होती हैं, इसलिए यह प्राथमिक सेल है और इसे पुनः आवेशित नहीं किया जा सकता।

Question 24Electrochemistry

किसी विलयन में Cu²⁺ तथा Ag⁺ दोनों आयन उपस्थित हैं और विद्युत-अपघटन किया जाता है। E°(Cu²⁺/Cu) = +0.34 V तथा E°(Ag⁺/Ag) = +0.80 V है। कैथोड पर पहले कौन-सा आयन अपचयित होगा?

  1. दोनों एक साथ, क्योंकि सांद्रताएँ समान हैं
  2. Cu²⁺, क्योंकि इसका परमाणु क्रमांक कम है
  3. Cu²⁺, क्योंकि इसका विभव कम धनात्मक है
  4. Ag⁺, क्योंकि इसका अपचयन विभव अधिक धनात्मक है Correct answer.

Explanation

कैथोड पर वही स्पीशीज़ पहले अपचयित होती है जिसका अपचयन विभव अधिक धनात्मक होता है, क्योंकि उसका अपचयन अधिक सरल होता है। यहाँ Ag⁺ का विभव (+0.80 V) Cu²⁺ के विभव (+0.34 V) से अधिक है, अतः सिल्वर पहले निक्षेपित होगा।

Question 25Electrochemistry

किसी सेल में निम्नलिखित अभिक्रिया होती है: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s)। E°(Fe³⁺/Fe²⁺) = +0.77 V तथा E°(I₂/I⁻) = +0.54 V है। इस अभिक्रिया के लिए E°(सेल) क्या होगा?

  1. +1.31 V
  2. +0.23 V Correct answer.
  3. −0.23 V
  4. +0.54 V

Explanation

यहाँ Fe³⁺ का अपचयन कैथोड पर तथा I⁻ का ऑक्सीकरण ऐनोड पर होता है, अतः E°(सेल) = E°(कैथोड) − E°(ऐनोड) = 0.77 − 0.54 = +0.23 V। धनात्मक मान दर्शाता है कि अभिक्रिया मानक परिस्थितियों में स्वतःप्रवर्तित है।

Question 26Electrochemistry

किसी सेल का E°(सेल) = +1.10 V है तथा अभिक्रिया में 2 इलेक्ट्रॉन स्थानांतरित होते हैं। 25 °C पर ΔG° का मान लगभग कितना होगा? (F = 96500 C mol⁻¹)

  1. +212.3 kJ mol⁻¹
  2. −106.2 kJ mol⁻¹
  3. −1.10 kJ mol⁻¹
  4. −212.3 kJ mol⁻¹ Correct answer.

Explanation

संबंध ΔG° = −nFE° का उपयोग करने पर ΔG° = −2 × 96500 × 1.10 = −212300 J mol⁻¹, अर्थात लगभग −212.3 kJ mol⁻¹। ऋणात्मक मान बताता है कि अभिक्रिया ऊष्मागतिकीय रूप से स्वतःप्रवर्तित है।

Question 27Electrochemistry

सीसा-संचायक सेल के डिस्चार्ज के दौरान कौन-सा परिवर्तन सही है?

  1. ऐनोड पर PbSO₄ का Pb बनना तथा कैथोड पर PbSO₄ का PbO₂ बनना
  2. विद्युत-अपघट्य में H₂SO₄ की सांद्रता बढ़ना
  3. ऐनोड पर Pb का PbSO₄ बनना तथा कैथोड पर PbO₂ का PbSO₄ बनना Correct answer.
  4. केवल कैथोड पर Pb का PbO₂ बनना

Explanation

डिस्चार्ज के समय ऐनोड पर Pb का ऑक्सीकरण होकर PbSO₄ बनता है तथा कैथोड पर PbO₂ का अपचयन होकर भी PbSO₄ बनता है। दोनों इलेक्ट्रोडों पर अविलेय PbSO₄ जमा होने से विद्युत-अपघट्य में H₂SO₄ की सांद्रता घटती है।

Question 28Electrochemistry

किसी सेल में विद्युत-अपघट्य की सांद्रता बढ़ाने पर विद्युत-चालकता तथा मोलर चालकता पर क्या प्रभाव पड़ता है?

  1. विद्युत-चालकता घटती है तथा मोलर चालकता बढ़ती है
  2. विद्युत-चालकता बढ़ती है परंतु मोलर चालकता सामान्यतः घटती है Correct answer.
  3. दोनों पर कोई प्रभाव नहीं पड़ता
  4. विद्युत-चालकता बढ़ती है तथा मोलर चालकता भी बढ़ती है

Explanation

सांद्रता बढ़ाने पर प्रति इकाई आयतन में आयनों की संख्या बढ़ती है, अतः विद्युत-चालकता बढ़ती है। परंतु प्रबल विद्युत-अपघट्यों में आयन-आयन अंतर्क्रियाएँ बढ़ने से प्रति मोल आयनों की गतिशीलता घटती है, इसलिए मोलर चालकता सामान्यतः घटती है।

Question 29Electrochemistry

किसी विद्युत-अपघट्य के विलयन से 0.5 A की धारा 2 घंटे तक प्रवाहित की जाती है। परिपथ से गुजरे आवेश का मान कितना होगा?

  1. 1000 C
  2. 7200 C
  3. 1800 C
  4. 3600 C Correct answer.

Explanation

आवेश Q = I × t, जहाँ समय सेकंड में लिया जाता है। यहाँ I = 0.5 A तथा t = 2 × 3600 = 7200 s, अतः Q = 0.5 × 7200 = 3600 C प्राप्त होता है।

Question 30Electrochemistry

किसी अभिक्रिया में 0.1 mol Cu²⁺ को Cu में अपचयित करने के लिए आवश्यक आवेश तथा उससे निक्षेपित ताँबे का द्रव्यमान क्रमशः क्या होंगे? (F = 96500 C mol⁻¹, Cu = 63.5 g mol⁻¹)

  1. 19300 C तथा 3.175 g
  2. 9650 C तथा 6.35 g
  3. 19300 C तथा 6.35 g Correct answer.
  4. 9650 C तथा 3.175 g

Explanation

Cu²⁺ + 2e⁻ → Cu के अनुसार प्रति मोल ताँबे के लिए 2F आवेश चाहिए, अतः 0.1 mol के लिए Q = 0.1 × 2 × 96500 = 19300 C। निक्षेपित द्रव्यमान m = 0.1 × 63.5 = 6.35 g होगा।

Question 31Chemical Kinetics

For a reaction A → products, the rate law is found to be rate = k[A]². What is the overall order of this reaction?

  1. First order
  2. Third order
  3. Second order Correct answer.
  4. Zero order

Explanation

The overall order is the sum of the exponents of the concentration terms in the experimentally determined rate law. Here the exponent on [A] is 2, so the sum is 2 and the reaction is second order overall.

Question 32Chemical Kinetics

The rate constant of a first-order reaction has the unit:

  1. L mol⁻¹ s⁻¹
  2. mol L⁻¹ s⁻¹
  3. L² mol⁻² s⁻¹
  4. s⁻¹ Correct answer.

Explanation

For a first-order reaction rate = k[A], so k = rate/[A] and its unit is (mol L⁻¹ s⁻¹)/(mol L⁻¹) = s⁻¹. The unit of k therefore reveals the overall order of the reaction.

Question 33Chemical Kinetics

In the Arrhenius equation k = A e^(−Ea/RT), what does the symbol A represent?

  1. The frequency or pre-exponential factor Correct answer.
  2. The equilibrium constant
  3. The half-life of the reaction
  4. The activation energy of the reaction

Explanation

In the Arrhenius equation A is the pre-exponential or frequency factor, related to the number of collisions with the correct orientation. The exponential term e^(−Ea/RT) then gives the fraction of collisions with enough energy to react.

Question 34Chemical Kinetics

For a zero-order reaction, how does the half-life depend on the initial concentration a?

  1. t½ is independent of a
  2. t½ is inversely proportional to a
  3. t½ is directly proportional to a Correct answer.
  4. t½ is proportional to a²

Explanation

For a zero-order reaction the integrated rate law gives t½ = a/(2k), so the half-life is directly proportional to the initial concentration. This contrasts with first order, where t½ = 0.693/k is independent of concentration.

Question 35Chemical Kinetics

A first-order reaction has a half-life of 20 minutes. Starting from 0.80 mol L⁻¹ of reactant, what concentration remains after 60 minutes?

  1. 0.20 mol L⁻¹
  2. 0.10 mol L⁻¹ Correct answer.
  3. 0.05 mol L⁻¹
  4. 0.40 mol L⁻¹

Explanation

Sixty minutes is three half-lives of 20 minutes each. Each half-life halves the concentration: 0.80 → 0.40 → 0.20 → 0.10 mol L⁻¹, so 0.10 mol L⁻¹ remains after three half-lives.

Question 36Chemical Kinetics

For the reaction 2N₂O₅ → 4NO₂ + O₂, the rate of disappearance of N₂O₅ is 4.0 × 10⁻³ mol L⁻¹ s⁻¹. What is the rate of formation of NO₂?

  1. 4.0 × 10⁻³ mol L⁻¹ s⁻¹
  2. 2.0 × 10⁻³ mol L⁻¹ s⁻¹
  3. 1.0 × 10⁻³ mol L⁻¹ s⁻¹
  4. 8.0 × 10⁻³ mol L⁻¹ s⁻¹ Correct answer.

Explanation

From the stoichiometry, 2 mol of N₂O₅ give 4 mol of NO₂, so the rate of formation of NO₂ is twice the rate of disappearance of N₂O₅. Thus 2 × 4.0 × 10⁻³ = 8.0 × 10⁻³ mol L⁻¹ s⁻¹.

Question 37Chemical Kinetics

Which statement correctly describes the effect of a catalyst on a reaction?

  1. It increases the enthalpy change of the reaction
  2. It increases the activation energy so more collisions occur
  3. It shifts the equilibrium position towards the products
  4. It lowers the activation energy without being consumed overall Correct answer.

Explanation

A catalyst provides an alternative pathway of lower activation energy, increasing the fraction of successful collisions, and it is regenerated at the end of the reaction. It speeds both forward and reverse reactions equally, so the equilibrium position and enthalpy change are unaffected.

Question 38Chemical Kinetics

The rate of a reaction becomes four times when the concentration of reactant A is doubled, while the concentration of B is kept constant. What is the order with respect to A?

  1. Second order Correct answer.
  2. Zero order
  3. Half order
  4. First order

Explanation

If doubling [A] multiplies the rate by 2ⁿ, then 2ⁿ = 4 gives n = 2. The reaction is therefore second order with respect to A, since rate ∝ [A]².

Question 39Chemical Kinetics

For a first-order reaction, a graph of log₁₀(reactant concentration) against time is:

  1. A straight line with a positive slope
  2. A parabola opening upward
  3. A straight line with a negative slope Correct answer.
  4. A horizontal straight line

Explanation

The integrated first-order rate law is log₁₀[A] = log₁₀[A]₀ − (k/2.303)t, which is linear in t with slope −k/2.303. The negative slope reflects the steady exponential decay of the reactant.

Question 40Chemical Kinetics

According to collision theory, which factor does NOT increase the rate of a gaseous reaction between two reactants?

  1. Adding a suitable catalyst
  2. Increasing the pressure of the gaseous mixture
  3. Increasing the activation energy Correct answer.
  4. Increasing the temperature

Explanation

The rate depends on the frequency of effective collisions, which rises with temperature, pressure, and the presence of a catalyst. Raising the activation energy reduces the fraction of molecules with sufficient energy, so the rate decreases rather than increases.

Question 41Chemical Kinetics

The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Using the Arrhenius equation, the activation energy is closest to: (R = 8.314 J K⁻¹ mol⁻¹, ln 2 = 0.693)

  1. About 108 kJ mol⁻¹
  2. About 5.4 kJ mol⁻¹
  3. About 54 kJ mol⁻¹ Correct answer.
  4. About 27 kJ mol⁻¹

Explanation

Using ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), we get 0.693 = (Ea/8.314)(1/300 − 1/310). The bracket equals 1.075 × 10⁻⁴, so Ea ≈ 0.693 × 8.314 / 1.075 × 10⁻⁴ ≈ 5.4 × 10⁴ J mol⁻¹, i.e. about 54 kJ mol⁻¹.

Question 42Chemical Kinetics

For the reaction 2A + B → products, the rate law is rate = k[A][B]. If the concentration of A is tripled and that of B is halved, the rate becomes:

  1. 3 times the original rate
  2. 1.5 times the original rate Correct answer.
  3. 6 times the original rate
  4. 0.67 times the original rate

Explanation

Substituting the new concentrations into rate = k[A][B] gives k(3[A])(0.5[B]) = 1.5 k[A][B]. The rate therefore increases by a factor of 1.5, the product of the two concentration factors.

Question 43Chemical Kinetics

Which of the following is a valid unit for the rate of a chemical reaction in solution?

  1. mol L⁻¹ s⁻¹ Correct answer.
  2. mol s⁻¹ L
  3. L mol⁻¹ s⁻¹
  4. mol L⁻¹

Explanation

Reaction rate is the change in concentration per unit time, so its unit is concentration divided by time, that is mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹). The unit mol L⁻¹ s⁻¹ also happens to be the unit of k for a zero-order reaction.

Question 44Chemical Kinetics

A reaction has the rate law rate = k[A]²[B]⁰. Doubling the concentration of B while keeping [A] constant will:

  1. Have no effect on the rate Correct answer.
  2. Double the rate
  3. Quadruple the rate
  4. Halve the rate

Explanation

The exponent of [B] in the rate law is zero, so [B]⁰ = 1 and the rate is independent of the concentration of B. Changing [B] therefore leaves the rate unchanged, even though B appears in the overall equation.

Question 45Chemical Kinetics

For a reaction, the rate constant is 2.0 × 10⁻³ s⁻¹ at 300 K and 8.0 × 10⁻³ s⁻¹ at 320 K. What is the order of the reaction, and what does the temperature dependence show?

  1. First order, since k decreases with temperature
  2. First order, since k has unit s⁻¹ and increases with temperature Correct answer.
  3. Second order, since k quadruples over 20 K
  4. Zero order, since the rate is constant

Explanation

The unit s⁻¹ identifies a first-order reaction, because k = rate/[A] carries units of reciprocal time. The rate constant increases with temperature in accordance with the Arrhenius equation, confirming the expected positive activation energy.

Question 46Chemical Kinetics

In the Arrhenius plot of ln k against 1/T, the slope of the straight line equals:

  1. −Ea
  2. A
  3. −Ea/R Correct answer.
  4. Ea/R

Explanation

Taking logarithms of k = A e^(−Ea/RT) gives ln k = ln A − (Ea/R)(1/T). Comparing with y = mx + c shows that the slope m is −Ea/R and the intercept is ln A.

Question 47The d- and f-Block Elements and Coordination Compounds

In the complex [Cr(NH3)4Cl2]+, the chromium ion is in which oxidation state?

  1. +3 Correct answer.
  2. +1
  3. +2
  4. +4

Explanation

Each NH3 is neutral and each chloride carries a -1 charge. The overall charge is +1, so x + 0 + 2(-1) = +1, giving x = +3. Chromium(III) is the common stable state here.

Question 48The d- and f-Block Elements and Coordination Compounds

Which statement correctly describes the electronic configuration of the Cr atom in its ground state?

  1. [Ar] 3d5 4s1 Correct answer.
  2. [Ar] 3d6
  3. [Ar] 3d3 4s2 4p1
  4. [Ar] 3d4 4s2

Explanation

Half-filled subshells give extra exchange stability, so chromium adopts [Ar] 3d5 4s1 rather than the expected [Ar] 3d4 4s2. This is a standard exception to the aufbau order among the 3d elements.

Question 49The d- and f-Block Elements and Coordination Compounds

The lanthanoid contraction is best explained by which factor?

  1. Increasing shielding by the 4f electrons as the series is crossed
  2. Loss of the 6s electrons progressively across the series
  3. A decrease in the principal quantum number of the valence shell
  4. Poor shielding of the 4f electrons, so effective nuclear charge rises across the series Correct answer.

Explanation

The 4f orbitals are diffuse and shield the outer electrons poorly. As nuclear charge increases across the lanthanoids, the outer electrons feel a greater effective nuclear charge and the radius contracts steadily.

Question 50The d- and f-Block Elements and Coordination Compounds

Which property is characteristic of transition metals and arises from their partly filled d orbitals?

  1. Inability to act as catalysts
  2. Strictly fixed coordination number of four
  3. Complete absence of variable oxidation states
  4. Formation of coloured ions and paramagnetism Correct answer.

Explanation

Partly filled d orbitals allow d-d transitions that absorb visible light, giving colour, and leave unpaired electrons that give paramagnetism. These features are hallmarks of transition-metal chemistry.