 Problem 1:
 H_{2} (g) + 1/2 S_{2} (g) = H_{2}S (g), KP_{1}
 2H_{2} (g) + S_{2} (g) = 2H_{2}S (g), KP_{2}

Show that, KP_{2} = (KP_{1})^{2}
 Answer:

For the first reaction,Î”G_{1}^{0} =  RT ln K_{P1}
For the second reaction, Î”G_{2}^{0} =  RT ln K_{P2}
Since, Î”G_{2}^{0} = 2Î”G_{1}^{0},
therefore, it follows that,
 RT ln K_{P2} =  2RT ln K_{P1}
∴ K_{P2} = (K_{P1})^{2} 
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,
Î”G^{0} = Î”H^{0} + T[d(Î”G^{0})/dT]_{P}
Î”G^{0} = Î”H^{0} + T[d(Î”G^{0})/dT]_{P}
Zero superscript is indicating the stranded values.
or,  (Î”H^{0}/T^{2} ) = (Î”G^{0}/T^{2} ) + 1/T[d(Î”G^{0})/dT]_{P}
or,  (Î”H^{0}/T^{2} ) = [d/dT(Î”G^{0}/T)]_{P}
Again Vant Hoff isotherm is,
 RT lnK_{P} = Î”G^{0}
or,  R lnK_{P} = Î”G^{0}/T
Differentiating with respect to T at constant P,
 R [dlnKP/dT]_{P} = [d/dT(Î”G^{0}/T)]_{P}
Comparing the above two equation we have,
dlnK_{P}/dT = (Î”H^{0}/T^{2} 

This is the Van't Hoff Equation isochore.

Greater the value of Î”H^{0}, the faster the equilibrium constant(K_{P})changes with temperature(T), Î”H^{0} should remain constant for the linear plot of logK_{P} 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:
 Equilibrium constant remains the same when P is increased.
 The reaction is exothermic, hence Î”H = () ve so the equilibrium constant is increased with decreases of temperature.
 Equilibrium constant remains the same though a catalyst is added. Î”G_{0} =  RT lnK_{a} but Î”G_{0} 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 (K_{a}) remains unchanged.
 Problem 4:

For a reaction,2A + B ⇆ 2C, Î”G^{0}(500 K) = 2 KJ mol^{1}. Find the K_{P} at 500 K for the reaction A + ½B ⇆ C.
 Answer:
Î”G^{0} (500 K) for the reaction,
A + ½B ⇆ C
= (2 KJ mol^{1})/2
= 1 KJ mol^{1}
The relation is, Î”G^{0} =  RT lnK_{P}
or, 1 =  (8.31 × 103) × (500) lnK_{P}
or, lnK_{P} = 1/(8.31 × 0.5)
= 0.2406
∴ K_{P} = 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 (Î”Î³ ≠ 0), if an inert gas is added at constant P (Î”Î³ ≠ 0), and any of the reacting components are added a depleted.

For example, N_{2} + 3 H_{2} ⇆ 2NH_{3}

Equilibrium yield of NH_{3} 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 Î”H (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 Î”H remains the same whether a catalyst is used or not.

Therefore the statement is correct.