Chemical equilibrium questions answers

  • Problem 1:
  1. H2 (g) + 1/2 S2 (g) = H2S (g), KP1
  2. 2H2 (g) + S2 (g) = 2H2S (g), KP2
    Show that, KP2 = (KP1)2
  • Answer:
    For the first reaction,ΔG10 = - RT ln KP1
    For the second reaction, ΔG20 = - RT ln KP2
    Since, ΔG20 = 2ΔG10,
    therefore, it follows that,
    - RT ln KP2 = - 2RT ln KP1
∴ KP2 = (KP1)2
Chemical Equilibrium Questions and Answers
Equilibrium Constant
  • Problem 2:
    Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. Under What condition do you expect a linear relationship between logk and 1/T ?
  • Answer:
The Gibbs - Helmholtz equation is,
ΔG0 = ΔH0 + T[d(ΔG0)/dT]P
Zero superscript is indicating the stranded values.
or, - (ΔH0/T2 ) = -(ΔG0/T2 ) + 1/T[d(ΔG0)/dT]P
or, - (ΔH0/T2 ) = [d/dT(ΔG0/T)]P
Again Vant Hoff isotherm is,
- RT lnKP = ΔG0
or, - R lnKP = ΔG0/T
Differentiating with respect to T at constant P,
- R [dlnKP/dT]P = [d/dT(ΔG0/T)]P
Comparing the above two equation we have,
dlnKP/dT = (ΔH0/T2
    Greater the value of ΔH0, the faster the equilibrium constant(KP)changes with temperature(T), ΔH0 should remain constant for the linear plot of logKP vs 1/T.
  • Problem 3
    How does the equilibrium constant for a reaction, 2A + 3B 4C + Heat, Change when (i) pressure is increasing (ii) Temperature is decreasing (iii) a catalyst added?
  • Answer:
  1. Equilibrium constant remains the same when P is increased.
  2. The reaction is exothermic, hence ΔH = (-) ve so the equilibrium constant is increased with decreases of temperature.
  3. Equilibrium constant remains the same though a catalyst is added. ΔG0 = - RT lnKa but ΔG0 is not changed due to the addition of catalyst as the latter does not participate in the reaction. The initial state and the final state of a chemical reaction remains the same whether the catalyst is added or not. Hence the equilibrium constant (Ka) remains unchanged.
  • Problem 4:
    For a reaction,2A + B 2C, ΔG0(500 K) = 2 KJ mol-1. Find the KP at 500 K for the reaction A + ½B C.
  • Answer:
ΔG0 (500 K) for the reaction,
A + ½B C 
= (2 KJ mol-1)/2
= 1 KJ mol-1
The relation is, ΔG0 = - RT lnKP
or, 1 = - (8.31 × 10-3) × (500) lnKP
or, lnKP = 1/(8.31 × 0.5)
= 0.2406
KP = 1.27
  • Problem 5:
    Justify or criticize the following: "The Equilibrium yield of products can not change if the equilibrium constant is kept fixed."
  • Answer:
    This statement is not correct.
    Equilibrium yield of the product is changed if pressure is changed (Δγ ≠ 0), if an inert gas is added at constant P (Δγ ≠ 0), and any of the reacting components are added a depleted.
    For example, N2 + 3 H2 2NH3
    Equilibrium yield of NH3 is increased if P is increased though equilibrium constant kept fixed.
  • Problem 6:
    Justify or criticize the following: Heat of reaction is the same whether a catalyst is used or not.
  • Answer:
    H is a state function hence ΔH (heat of a reaction) does not change if initial state and final state of a Process is same. A catalyst cannot change the initial and final state of a chemical reaction, hence ΔH remains the same whether a catalyst is used or not.
    Therefore the statement is correct.

Chemical Equilibrium Questions and Answers: Law of Mass Action and Van't Hoff Equation.

Inorganic Chemistry

[Inorganic chemistry][column1]

Contact Form


Email *

Message *

Powered by Blogger.