### Chemical Equilibrium Questions and Answers

Given,

(i) H_{2} (g) + ^{1}/_{2} S_{2} (g) = H_{2}S (g), K _{P1} |

(ii) 2H_{2} (g) + S_{2} (g) = 2H_{2}S (g), K _{P2} |

Show that,

**K**_{P2}= (K_{P1})^{2}
For the first reaction, Î”G

_{1}^{0}= - RT lnK_{P1}
For the second reaction, Î”G

_{2}^{0}= - RT lnK_{P2}
Since, Î”G

_{2}^{0}= 2Î”G_{1}^{0}, therefore, it follows that,
- RT lnK

_{P2}= - 2RT lnK_{P1}∴ K_{P2} = (K_{P1})^{2} |

Equilibrium Constant |

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 ?

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 remains constant for the linear plot of^{0}**logK**vs_{P}**1/T**.
How does the equilibrium constant for a reaction,
2A + 3B ⇆ 4C + Heat, Change when (i) pressure is increases (ii) Temperature is decreases (iii) a catalyst added?

__Equilibrium constant__remains same when P is increases .- The reaction is exothermic, hence
**Î”H = (-)ve**so the equilibrium constant is increased with decreases of temperature. - Equilibrium constant remains same though a catalyst is added.

**Î”G**but

^{0}= - RT lnK_{a}**Î”G**is not changed due to addition of catalyst as the latter does not participate in the reaction. The initial state and the final state of a chemical reaction remains same weather the catalyst is added or not.

^{0}
Hence the equilibrium constant (

**K**) remains unchanged._{a}
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.
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
Justify or criticize the following:
"The Equilibrium yield of products can not change if the equilibrium constant is kept fixed."

Equilibrium yield of the product is changed if pressure is changed (Î”Î³ ≠ 0), if inert gas is added at constant

**P**(Î”Î³ ≠ 0), and any of the reacting component is 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.
Justify or criticize the following: Heat of reaction is the same weather a catalyst is used or not.

**H**is a state function hence

**Î”H**(heat of a reaction) does no change if initial state and final state of a Process is same.

A catalyst can not change the initial and final state of a chemical reaction, hence

**Î”H**remains same weather a catalyst is used or not.
Therefore the statement is correct.