The plot shows the variation of -ln⁡K_P Versus temperature for the two reactions. Ms+1/2O_2g→MO(s) and Cs+1/2O_2g→CO(s) Identify the correct statement:

Correct answer: (C) At T<1200 K, the reaction MOs+Cs→Ms+CO(g) is spontaneous. ΔG=ΔG^0 +RTlnQ At Equilibrium ΔG=0 & Q=K ∆G^o=-RTln⁡Kp ∴-ln⁡K_p=(∆G^o)/RT ∆G^o∝ Temperature. At high temperature ∆G^o value is very high so non-feasible or non-spontaneous process takes place.

Q19 / 90 · JEE Main 2016 (09 Apr Online) · Chemistry

The plot shows the variation of $- \ln K_{P}$ Versus temperature for the two reactions.

$M (s) + \frac{1}{2} O_{2} (g) \to M O (s)$ and

$C (s) + \frac{1}{2} O_{2} (g) \to C O (s)$

Identify the correct statement:

A
At $T < 1200 K$ , oxidation of carbon is unfavourable
B
Oxidation of carbon is favourable at all temperatures
Correct answer
C
At $T < 1200 K$ , the reaction $M O (s) + C (s) \to M (s) + C O (g)$ is spontaneous
D
At $T > 1200 K$ , carbon will reduce ${MO}_{(s)} to M_{(s)}$ .

Solution

Δ G = Δ G^{0} + R T \ln Q

At Equilibrium

Δ G = 0

& Q = K

∆ G^{o} = - R T \ln K p

∴- \ln K_{p} = \frac{∆ G^{o}}{R T}

∆ G^{o} \propto

Temperature. At high temperature

∆ G^{o}

value is very high so non-feasible or non-spontaneous process takes place.

Hence, the correct answer is option: 'C'.

The plot shows the variation of -ln⁡KP Versus temperature for the two reactions.Ms+12O2g→MO(s) andCs+12O2g→CO(s)Identify the correct statement:

Solution

Relevant Questions

The plot shows the variation of $- \ln K_{P}$ Versus temperature for the two reactions.

$M (s) + \frac{1}{2} O_{2} (g) \to M O (s)$ and

$C (s) + \frac{1}{2} O_{2} (g) \to C O (s)$

Identify the correct statement: