- 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

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**, the faster the equilibrium constant(

^{0}**K**)changes with temperature(T),

_{P}**Î”H**should remain constant for the linear plot of

^{0}**logK**vs

_{P}**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**but_{0}= - RT lnK_{a}**Î”G**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 (_{0}**K**) remains unchanged._{a}

- 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 × 10-3) × (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.__