I85PREP
⌘ K
WBCHSE HS Chemistry study materials
WBCHSE HS Chemistry study resources

WBCHSE HS Chemistry practice questions

One question at a time. Build your WBCHSE HS 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 1Atomic Structure and Chemical Bonding

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in the n = 3 orbit is:

  1. 2h/π
  2. 3h/2π Correct answer.
  3. h/2π
  4. 9h/2π

Explanation

Bohr's quantisation condition states that angular momentum mvr equals nh/2π, where n is the principal quantum number. For the third orbit n = 3, so the angular momentum is 3h/2π.

Question 2Atomic Structure and Chemical Bonding

The maximum number of electrons that can be accommodated in the subshell with azimuthal quantum number l = 2 is:

  1. 10 Correct answer.
  2. 14
  3. 2
  4. 6

Explanation

A subshell with l = 2 is a d subshell, which has (2l + 1) = 5 orbitals. Each orbital holds two electrons with opposite spins, giving a maximum of 10 electrons.

Question 3Atomic Structure and Chemical Bonding

Which of the following species is isoelectronic with the neon atom?

  1. Ca²⁺
  2. Cl⁻
  3. S²⁻
  4. Na⁺ Correct answer.

Explanation

Neon has 10 electrons. The Na⁺ ion has 11 − 1 = 10 electrons, so it is isoelectronic with neon. The other ions contain 18 electrons each.

Question 4Atomic Structure and Chemical Bonding

The shape of the ammonia molecule, NH₃, according to VSEPR theory, is:

  1. trigonal pyramidal Correct answer.
  2. tetrahedral
  3. T-shaped
  4. trigonal planar

Explanation

Nitrogen in NH₃ has three bond pairs and one lone pair, giving a steric number of four. The electron-pair geometry is tetrahedral, but the molecular shape described by the atoms is trigonal pyramidal.

Question 5Atomic Structure and Chemical Bonding

Which bond is formed by the head-on overlap of atomic orbitals along the internuclear axis?

  1. σ bond Correct answer.
  2. δ bond
  3. Hydrogen bond
  4. π bond

Explanation

A σ bond results from axial or head-on overlap of orbitals, giving maximum electron density along the internuclear axis. π bonds arise from sideways overlap of parallel p orbitals.

Question 6Atomic Structure and Chemical Bonding

The number of unpaired electrons in the ground-state configuration of the Fe³⁺ ion (Z = 26) is:

  1. 6
  2. 4
  3. 3
  4. 5 Correct answer.

Explanation

Neutral Fe is [Ar]3d⁶4s². Removing three electrons gives Fe³⁺ as [Ar]3d⁵, and by Hund's rule the five d electrons occupy all five orbitals singly. Hence there are five unpaired electrons.

Question 7Atomic Structure and Chemical Bonding

The hybridisation of the carbon atom in the ethyne (acetylene) molecule is:

  1. sp²
  2. sp³
  3. sp³d
  4. sp Correct answer.

Explanation

Each carbon in ethyne forms two σ bonds (one to hydrogen, one to the other carbon) and two π bonds. Two σ bonds require two hybrid orbitals, so the carbon is sp hybridised with linear geometry.

Question 8Atomic Structure and Chemical Bonding

Which of the following sets of quantum numbers is not permitted for an electron in an atom?

  1. n = 1, l = 1, m = 0 Correct answer.
  2. n = 3, l = 2, m = −2
  3. n = 4, l = 0, m = 0
  4. n = 2, l = 1, m = 0

Explanation

The azimuthal quantum number l can take values from 0 up to (n − 1). For n = 1 only l = 0 is allowed, so the set n = 1, l = 1 is forbidden regardless of the value of m.

Question 9Atomic Structure and Chemical Bonding

The energy of an electron in the second Bohr orbit of the hydrogen atom is:

  1. −3.4 eV Correct answer.
  2. −13.6 eV
  3. −6.8 eV
  4. −1.51 eV

Explanation

Bohr's expression gives Eₙ = −13.6/n² eV for hydrogen. Substituting n = 2 gives −13.6/4 = −3.4 eV, the energy of the first excited state.

Question 10Atomic Structure and Chemical Bonding

The de Broglie wavelength of an electron accelerated through a potential difference of 100 V is approximately:

  1. 123 Å
  2. 0.123 Å
  3. 12.3 Å
  4. 1.23 Å Correct answer.

Explanation

For an electron accelerated through V volts, λ ≈ √(150/V) Å. Substituting V = 100 V gives λ ≈ √1.5 ≈ 1.23 Å, a value typical of electron diffraction experiments.

Question 11Atomic Structure and Chemical Bonding

The incorrect statement about the shapes of molecules according to VSEPR theory is that:

  1. a molecule with two bond pairs and three lone pairs is linear
  2. a molecule with three bond pairs and one lone pair is trigonal planar Correct answer.
  3. lone pairs occupy more space than bond pairs and reduce bond angles
  4. a molecule with four bond pairs and no lone pairs is tetrahedral

Explanation

In a molecule with three bond pairs and one lone pair, the electron-pair geometry is tetrahedral, but the molecular shape is trigonal pyramidal, as in NH₃. The bond angle is also compressed below 109.5°, so calling it trigonal planar is incorrect.

Question 12Atomic Structure and Chemical Bonding

The dipole moment of carbon dioxide is zero because:

  1. the molecule has a bent shape with equal bond angles
  2. carbon and oxygen have identical electronegativities
  3. the two C=O bond dipoles are equal and cancel in the linear geometry Correct answer.
  4. the C=O bonds are purely ionic

Explanation

Each C=O bond is polar, but the molecule is linear with the two bond dipoles pointing in exactly opposite directions. They are equal in magnitude and cancel, so the net dipole moment is zero.

Question 13Atomic Structure and Chemical Bonding

The bond order of the O₂⁻ ion according to molecular orbital theory is:

  1. 2.0
  2. 1.0
  3. 2.5
  4. 1.5 Correct answer.

Explanation

O₂⁻ has 17 electrons. Its configuration gives bonding electrons minus antibonding electrons equal to 3, and bond order equals 3/2 = 1.5. This places it between O₂ (2.0) and O₂²⁻ (1.0).

Question 14Atomic Structure and Chemical Bonding

Which of the following molecules is paramagnetic according to molecular orbital theory?

  1. F₂
  2. CO
  3. O₂ Correct answer.
  4. N₂

Explanation

The MO configuration of O₂ has two unpaired electrons in the degenerate π* antibonding orbitals, so it is paramagnetic. N₂, F₂ and CO are all diamagnetic with all electrons paired.

Question 15Atomic Structure and Chemical Bonding

In the formation of a hydrogen bond, the hydrogen atom is bonded to:

  1. another hydrogen atom by a covalent bond
  2. a highly electronegative atom such as F, O or N Correct answer.
  3. a metal atom through a metallic bond
  4. a large, low-electronegativity atom such as caesium

Explanation

Hydrogen bonding occurs when hydrogen is covalently bonded to a small, highly electronegative atom (F, O or N). The resulting strong dipole lets that hydrogen interact with a lone pair on a neighbouring electronegative atom.

Question 16Atomic Structure and Chemical Bonding

The bond angle in the water molecule (104.5°) is smaller than the tetrahedral angle mainly because:

  1. oxygen uses sp² hybridisation in water
  2. hydrogen atoms repel each other more than lone pairs do
  3. the two lone pairs on oxygen repel the bonding pairs more strongly Correct answer.
  4. the O–H bonds are ionic in character

Explanation

Oxygen in water is sp³ hybridised with two bond pairs and two lone pairs. Lone pairs occupy more space than bond pairs, so they push the O–H bonds closer together and reduce the angle below 109.5°.

Question 17Atomic Structure and Chemical Bonding

The formal charge on the central oxygen atom in the ozone molecule, O₃, when written with one double and one single bond, is:

  1. +2
  2. 0
  3. +1 Correct answer.
  4. −1

Explanation

For the central oxygen, formal charge equals valence electrons (6) minus non-bonding electrons (2) minus half the bonding electrons (3). This gives 6 − 2 − 3 = +1.

Question 18Atomic Structure and Chemical Bonding

The increasing order of bond length among the species N₂, N₂⁺ and N₂⁻ is:

  1. N₂⁺ < N₂⁻ < N₂
  2. N₂ < N₂⁺ < N₂⁻
  3. N₂⁻ < N₂ < N₂⁺
  4. N₂⁺ < N₂ < N₂⁻ Correct answer.

Explanation

Bond order for N₂ is 3, for N₂⁺ is 2.5 and for N₂⁻ is 2.5 as well, but the extra electron in N₂⁻ enters an antibonding orbital while removal in N₂⁺ takes an electron from a bonding orbital. The effective bond orders are 3, 2.5 and 2.0 respectively, so length increases as N₂⁺ < N₂ < N₂⁻.

Question 19Atomic Structure and Chemical Bonding

The number of radial nodes and angular nodes respectively for a 4d orbital are:

  1. 2 and 0
  2. 1 and 2 Correct answer.
  3. 2 and 1
  4. 0 and 2

Explanation

The number of angular nodes equals l, which is 2 for a d orbital. The number of radial nodes equals n − l − 1 = 4 − 2 − 1 = 1, so a 4d orbital has one radial and two angular nodes.

Question 20Atomic Structure and Chemical Bonding

The wavelength associated with the emission line corresponding to the transition n = 4 to n = 2 in the hydrogen spectrum can be calculated using the Rydberg equation. The value of 1/λ is:

  1. R/16
  2. R/4
  3. 5R/16
  4. 3R/16 Correct answer.

Explanation

The Rydberg equation gives 1/λ = R(1/n₁² − 1/n₂²). For the transition from n = 4 to n = 2, this becomes R(1/4 − 1/16) = R(3/16) = 3R/16, a line of the Balmer series.

Question 21Atomic Structure and Chemical Bonding

The correct order of hybridisation of the central atom in SF₄, SF₆ and PCl₅ is:

  1. sp³d, sp³, sp³d²
  2. sp³d, sp³d², sp³d Correct answer.
  3. sp³d², sp³d, sp³d
  4. sp³, sp³d, sp³d²

Explanation

SF₄ has five electron pairs (four bond pairs plus one lone pair) and is sp³d hybridised with a see-saw shape. SF₆ has six bond pairs and is sp³d², while PCl₅ has five bond pairs and is sp³d.

Question 22Atomic Structure and Chemical Bonding

The bond angle in H₂S is smaller than that in H₂O. The most appropriate explanation is that:

  1. the S–H bond is shorter than the O–H bond
  2. H₂S has a linear geometry with a bond angle of 180°
  3. sulfur uses sp² hybridisation while oxygen uses sp³
  4. sulfur is less electronegative, so bonding pairs lie farther from the central atom and repel less Correct answer.

Explanation

Sulfur is larger and less electronegative than oxygen, so the bonding electron pairs lie farther from the central atom and their mutual repulsion weakens. The lone-pair repulsion pushes the S–H bonds closer, giving about 92° compared with 104.5° in water.

Question 23Atomic Structure and Chemical Bonding

Which of the following pairs of species have the same bond order but different magnetic behaviour?

  1. O₂ and O₂⁺ Correct answer.
  2. CN⁻ and NO⁺
  3. N₂ and N₂⁺
  4. F₂ and O₂²⁻

Explanation

O₂ has bond order 2.0 and is paramagnetic with two unpaired π* electrons. Removing one electron gives O₂⁺ with bond order 2.5, so this pair differs in both bond order and magnetism, and the question is resolved by checking each pair: CN⁻ and NO⁺ are both 14-electron species with bond order 3 and are both diamagnetic. The intended comparison is between O₂ and O₂⁺, which must be rejected; the correct pair is CN⁻ and NO⁺, so this option is not the answer.

Question 24Atomic Structure and Chemical Bonding

The energy of the photon emitted when a hydrogen atom falls from n = 3 to n = 2 is closest to:

  1. 12.09 eV
  2. 1.89 eV Correct answer.
  3. 10.20 eV
  4. 3.40 eV

Explanation

Using E = 13.6(1/n₁² − 1/n₂²) eV with n₁ = 2 and n₂ = 3 gives 13.6(1/4 − 1/9) = 13.6 × 5/36 ≈ 1.89 eV. This is the red Hα line of the Balmer series.

Question 25States of Matter and Thermodynamics

At constant temperature, a fixed mass of ideal gas has its pressure tripled. According to Boyle's law, what happens to its volume?

  1. It remains unchanged.
  2. It becomes nine times the original volume.
  3. It becomes three times the original volume.
  4. It becomes one-third of the original volume. Correct answer.

Explanation

Boyle's law states that at constant temperature the product PV is constant for a fixed amount of gas. Tripling P therefore forces V to fall to one-third of its initial value. The relationship is inverse, not direct or quadratic.

Question 26States of Matter and Thermodynamics

Which statement correctly describes the conditions under which a real gas most closely approaches ideal behaviour?

  1. Low pressure and high temperature Correct answer.
  2. High pressure and high temperature
  3. Low pressure and low temperature
  4. High pressure and low temperature

Explanation

Ideal behaviour requires negligible molecular volume and negligible intermolecular attraction. Both conditions are favoured when molecules are far apart and moving fast, that is at low pressure and high temperature. Under high pressure or low temperature the corrections become significant.

Question 27States of Matter and Thermodynamics

Which of the following quantities is a state function?

  1. Work done by the system
  2. Enthalpy Correct answer.
  3. Heat absorbed by the system
  4. Path length of the process

Explanation

A state function depends only on the current state of the system, not on how that state was reached. Enthalpy, internal energy, entropy and Gibbs energy are state functions. Heat and work depend on the path followed between the same initial and final states.

Question 28States of Matter and Thermodynamics

For an isothermal reversible expansion of an ideal gas, which quantity is zero?

  1. Change in internal energy Correct answer.
  2. Change in entropy of the gas
  3. Work done by the gas
  4. Heat absorbed by the gas

Explanation

The internal energy of an ideal gas depends only on temperature. Since the process is isothermal, the change in internal energy is zero, so the heat absorbed equals the work done. Entropy still increases because the volume increases.

Question 29States of Matter and Thermodynamics

The compressibility factor Z of a gas is defined as PV/nRT. For an ideal gas at all pressures, Z equals:

  1. Greater than 1
  2. 1 Correct answer.
  3. Less than 1
  4. 0

Explanation

For one mole of ideal gas, PV = RT, so Z = PV/RT = 1 at every temperature and pressure. Deviations of Z from unity measure how far a real gas departs from ideality. Values above or below 1 indicate dominant repulsive or attractive effects respectively.

Question 30States of Matter and Thermodynamics

Which pair of thermodynamic quantities is related by the equation ΔG = ΔH − TΔS?

  1. Entropy change, heat and temperature
  2. Gibbs energy change, enthalpy change and entropy change Correct answer.
  3. Enthalpy change, internal energy change and pressure-volume work
  4. Internal energy change, work and heat

Explanation

The Gibbs energy equation links the change in Gibbs energy to enthalpy change, temperature and entropy change. It determines spontaneity at constant temperature and pressure. A negative ΔG corresponds to a spontaneous process.

Question 31States of Matter and Thermodynamics

Which of the following is an intensive property of a thermodynamic system?

  1. Internal energy
  2. Temperature Correct answer.
  3. Volume
  4. Mass

Explanation

Intensive properties are independent of the amount of substance present, and temperature is one of them. Mass, volume and internal energy all scale with the size of the system and are therefore extensive. Dividing an extensive property by amount of substance gives an intensive molar quantity.

Question 32States of Matter and Thermodynamics

A gas expands from 2 L to 8 L at a constant external pressure of 1 atm. What is the work done by the gas, expressed in litre-atmospheres?

  1. 6 L atm Correct answer.
  2. 16 L atm
  3. 4 L atm
  4. 10 L atm

Explanation

Pressure-volume work at constant external pressure is w = P_ext × ΔV. Here ΔV = 8 − 2 = 6 L, so the work done by the gas is 1 × 6 = 6 L atm. The sign convention for work done by the system is often taken as negative, but the magnitude is 6 L atm.

Question 33States of Matter and Thermodynamics

A sample of ideal gas occupies 300 mL at 27 °C. At constant pressure, what volume will it occupy at 127 °C?

  1. 400 mL Correct answer.
  2. 350 mL
  3. 129 mL
  4. 700 mL

Explanation

Charles's law gives V1/T1 = V2/T2 with temperatures in kelvin. Here T1 = 300 K and T2 = 400 K, so V2 = 300 × 400/300 = 400 mL. Using Celsius temperatures directly would give a wrong answer.

Question 34States of Matter and Thermodynamics

Equal volumes of hydrogen and oxygen are kept in identical containers at the same temperature and pressure. Which statement about the two gases is correct?

  1. They contain equal numbers of molecules. Correct answer.
  2. They have equal masses.
  3. They have equal densities.
  4. They have equal rates of diffusion.

Explanation

Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Because hydrogen and oxygen have different molar masses, their masses, densities and diffusion rates differ. The molecular count, however, is the same.

Question 35States of Matter and Thermodynamics

The critical temperature of a gas is the temperature above which:

  1. The gas cannot be liquefied however great the pressure Correct answer.
  2. The gas has zero compressibility factor
  3. The gas behaves exactly as an ideal gas
  4. The gas has maximum density

Explanation

Above the critical temperature, molecular kinetic energy is too high for attractive forces to hold molecules in a liquid state, regardless of applied pressure. Liquefaction therefore becomes impossible above this temperature. The critical pressure is the minimum pressure needed at that temperature.

Question 36States of Matter and Thermodynamics

For the reaction N2(g) + 3H2(g) → 2NH3(g), the value of Δn(gas) used in the relation ΔH = ΔU + ΔnRT is:

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

Explanation

Δn(gas) is the moles of gaseous products minus moles of gaseous reactants. Here 2 − (1 + 3) = −2. The negative value means the volume decreases and ΔH is smaller than ΔU by 2RT.

Question 37States of Matter and Thermodynamics

The standard enthalpy of formation of an element in its most stable standard state is taken as:

  1. Equal to its molar heat capacity
  2. Always negative
  3. Equal to its enthalpy of sublimation
  4. Zero at all temperatures by convention Correct answer.

Explanation

By convention, the standard enthalpy of formation of every element in its most stable form at standard conditions is set to zero. This provides a reference scale from which compound enthalpies are calculated. The convention applies to the element in its standard state, not to other allotropes.

Question 38States of Matter and Thermodynamics

A system absorbs 500 J of heat and does 200 J of work on the surroundings. What is the change in internal energy of the system?

  1. +300 J Correct answer.
  2. +700 J
  3. −700 J
  4. −300 J

Explanation

Using the convention ΔU = q + w, with q = +500 J for heat absorbed and w = −200 J for work done by the system, ΔU = 500 − 200 = +300 J. The internal energy rises because more heat enters than work leaves.

Question 39States of Matter and Thermodynamics

Which process has a negative entropy change for the system?

  1. Condensation of steam to liquid water at 100 °C Correct answer.
  2. Melting of ice at 0 °C
  3. Dissolution of ammonium chloride in water
  4. Expansion of a gas into a vacuum

Explanation

Entropy measures molecular disorder, and a gas is far more disordered than a liquid. When steam condenses, molecules lose translational freedom and entropy decreases. Melting, dissolution and expansion all increase disorder.

Question 40States of Matter and Thermodynamics

At what temperature does a reaction with ΔH = +40 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹ become spontaneous?

  1. Above 400 K Correct answer.
  2. Above 4 K
  3. At all temperatures
  4. Below 400 K

Explanation

Spontaneity requires ΔG = ΔH − TΔS < 0. Setting ΔH = TΔS gives T = 40000/100 = 400 K. Because both ΔH and ΔS are positive, the reaction is spontaneous only above this temperature.

Question 41States of Matter and Thermodynamics

According to the kinetic molecular theory of gases, the average kinetic energy of gas molecules is directly proportional to:

  1. Pressure
  2. Absolute temperature Correct answer.
  3. Volume
  4. Molar mass

Explanation

The kinetic theory gives average translational kinetic energy as (3/2)kT per molecule, which depends only on absolute temperature. Pressure and volume affect the number density but not the mean molecular energy. Molar mass affects molecular speed, not average kinetic energy.

Question 42States of Matter and Thermodynamics

Under the same conditions, which gas effuses fastest through a small pinhole?

  1. Methane (M = 16)
  2. Carbon dioxide (M = 44)
  3. Helium (M = 4) Correct answer.
  4. Nitrogen (M = 28)

Explanation

Graham's law states that the rate of effusion is inversely proportional to the square root of molar mass. Helium has the smallest molar mass among the listed gases, so it effuses fastest. The order of rates follows the inverse square roots of 4, 16, 28 and 44.

Question 43States of Matter and Thermodynamics

The van der Waals equation is (P + an²/V²)(V − nb) = nRT. Which term accounts for the attractive forces between molecules?

  1. an²/V² Correct answer.
  2. nRT
  3. P
  4. nb

Explanation

The correction an²/V² is added to the measured pressure because attractive forces reduce the pressure exerted on the walls. The constant b accounts for the finite volume occupied by molecules. The term nRT represents the ideal contribution to the equation.

Question 44States of Matter and Thermodynamics

For the isothermal reversible expansion of 1 mol of ideal gas from V to 2V at temperature T, the entropy change of the gas is:

  1. R ln 2 Correct answer.
  2. R ln(1/2)
  3. 2R ln 2
  4. Zero

Explanation

For an isothermal reversible process, ΔS = q_rev/T and q_rev = nRT ln(V2/V1). With n = 1 and V2/V1 = 2, this gives ΔS = R ln 2, a positive value because the volume increases.

Question 45States of Matter and Thermodynamics

A gas obeys the equation of state PV = nRT + nbP. Which statement about this gas is correct?

  1. It behaves ideally at all pressures.
  2. It has finite molecular volume but no intermolecular attraction. Correct answer.
  3. It cannot be liquefied at any temperature.
  4. It has intermolecular attraction but no molecular volume.

Explanation

Rearranging gives P(V − nb) = nRT, which is the van der Waals equation with the attraction term a set to zero. The constant b represents excluded volume, so the gas has finite molecular size but no attractive correction. Without attraction, liquefaction is not described by this model.

Question 46States of Matter and Thermodynamics

For an adiabatic reversible compression of an ideal gas, which combination of changes is correct?

  1. Temperature falls and entropy increases
  2. Temperature rises and entropy increases
  3. Temperature stays constant and entropy falls
  4. Temperature rises and entropy stays constant Correct answer.

Explanation

In a reversible adiabatic process q = 0, so entropy change is zero because ΔS = q_rev/T. Work done on the gas raises its internal energy and therefore its temperature. The compression follows PV^γ = constant.

Question 47States of Matter and Thermodynamics

The standard enthalpy of combustion of methane is −890 kJ mol⁻¹. What is the standard enthalpy of formation of methane, given that ΔHf°(CO2) = −394 kJ mol⁻¹ and ΔHf°(H2O, l) = −286 kJ mol⁻¹?

  1. +74 kJ mol⁻¹
  2. −210 kJ mol⁻¹
  3. −1570 kJ mol⁻¹
  4. −74 kJ mol⁻¹ Correct answer.

Explanation

For CH4 + 2O2 → CO2 + 2H2O, ΔHc° = [ΔHf°(CO2) + 2ΔHf°(H2O)] − ΔHf°(CH4). Substituting gives −890 = (−394 − 572) − ΔHf°(CH4), so ΔHf°(CH4) = −74 kJ mol⁻¹. The negative sign shows methane is more stable than its elements.

Question 48States of Matter and Thermodynamics

Two moles of an ideal gas expand isothermally and reversibly at 300 K from 10 L to 20 L. Taking R = 8.314 J K⁻¹ mol⁻¹, the work done by the gas is approximately:

  1. 4988 J
  2. 3457 J Correct answer.
  3. 2494 J
  4. 1729 J

Explanation

For isothermal reversible expansion, w = nRT ln(V2/V1) = 2 × 8.314 × 300 × ln 2. Since ln 2 ≈ 0.693, the work is about 3457 J. This equals the heat absorbed because ΔU is zero.

Question 49Chemical Equilibrium, Ionic Equilibrium and Redox

For the reversible reaction N2(g) + 3H2(g) ⇌ 2NH3(g), the equilibrium constant is expressed as Kc = [NH3]^2 / ([N2][H2]^3). If the reaction is rewritten as ½N2(g) + 3/2H2(g) ⇌ NH3(g), what is the new equilibrium constant in terms of the original Kc?

  1. 2Kc
  2. Kc/2
  3. Kc^2
  4. Kc^(1/2) Correct answer.

Explanation

When a balanced equation is multiplied by a factor n, the equilibrium constant is raised to the power n. Halving every coefficient means n = ½, so the new constant equals the square root of the original, Kc^(1/2). The exponent acts on Kc itself, not on the concentrations directly.

Question 50Chemical Equilibrium, Ionic Equilibrium and Redox

In the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), the equilibrium concentrations at a certain temperature are [PCl5] = 0.20 mol L⁻¹, [PCl3] = 0.10 mol L⁻¹ and [Cl2] = 0.10 mol L⁻¹. What is the value of Kc?

  1. 5.0
  2. 0.20
  3. 0.050 Correct answer.
  4. 0.020

Explanation

For this decomposition, Kc = [PCl3][Cl2]/[PCl5]. Substituting gives (0.10 × 0.10)/0.20 = 0.010/0.20 = 0.050. The product concentrations are multiplied because both appear in the balanced equation with coefficient one.