Calculate

(a) $$∆G^0$$ and

(b) the equilibrium constant for the formation of $$NO_2$$ from NO and $$O_2$$ at 298K

$$NO(g)+\dfrac{1}{2}O_{2}(g) ⇌ NO_{2}(g)$$

Where;

∆fG° $$(NO_2)$$ = 52.0 kJ/mol

∆fG° (NO) = 87.0 kJ/mol

∆fG° $$(O_2)$$ = 0 kJ/mol

Asked by Abhisek | 1 year ago |  135

##### Solution :-

(a) For the given reaction, we have

∆G° = ∆G°(Products) – ∆G°(Reactants)

∆G° = 52.0 – (87.0 + 0)

= -35.0 KJ mol-1

(b) We know that,

∆G° = RT log Kc

∆G° = 2.303 RT log Kc

$$K_{c}=\dfrac{-35.0\times 10^{-3}}{-2.303\times 8.314\times 298}\\ \\$$

$$=6.134\\ \\$$

$$K_{c}=antilog(6.134)\\ \\$$

$$=1.36\times 10^{6}$$

Therefore, the equilibrium constant for the given reaction Kc is

$$1.36\times 10^{6}$$

Answered by Pragya Singh | 1 year ago

### Related Questions

#### The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M.

The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2 . in which of these solutions precipitation will take place?

#### What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K?

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6).

#### What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18).