Chemical equilibrium problems solutions

Chemical Equilibrium Questions and Answers
Chemical equilibrium
Problem
H₂ (g) + ½ S₂ (g) = H₂S (g), KP₁
H₂ (g) + S₂ (g) = 2H₂S (g), KP₂
Show that, KP₂ = (KP₁)²
Solution
For the first reaction,
ΔG⁰₁ = - RT ln KP₁

For the second reaction,
ΔG⁰₂ = - RT ln KP₂
Since, ΔG⁰₂ = 2ΔG⁰₁
Therefore, it follows that - RT ln KP₂ = - 2RT ln KP₁

∴ Kp₂ = (Kp₁)²
Problem
    Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. What condition do you expect a linear relationship between log k and 1/T?
Solution
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 Van't 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 log Kp vs 1/T.
Problem
    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?
Solution

  1. Equilibrium constant remains the same when P is increased. The reaction is exothermic, hence ΔH = (-) ve so the equilibrium constant is increased with decreases of temperature.
  2. 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.
  3. 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
    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.
Solution
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
    Justify or criticize the following: "The equilibrium yield of products can not change if the equilibrium constant is kept fixed."
Solution
    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
    The equilibrium yield of NH₃ is increased if P is increased though equilibrium constant kept fixed.
Problem
    Justify or criticize the following: heat of reaction is the same whether a catalyst is used or not.
Solution
    H is a state function hence ΔH (heat of a reaction) does not change if the initial state and final state of a process are the 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.

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