The value of Kc for the following equilibrium is CaCO3(s) leftharpoons CaO(s) +CO2(g). Given Kp = 167 bar at 1073 K.

Question

The value of Kc for the following equilibrium is CaCO3(s) leftharpoons CaO(s) +CO2(g). Given Kp = 167 bar at 1073 K. (A) 1.896 mol L-1 (B) 4.38×10-4 mol L-1 (C) 6.3×104 mol L-1 (D) 6.626 mol L-1

Tardigrade Admin · Accepted Answer

Kp=Kc(RT)Δ n ; Δ n=1 Kp=167 bar, Kc=(167 bar/0.0821 L bar K-1 mol-1×1073 K) =1.896 mol L-1

Q. The value of $K_{c}$ for the following equilibrium is
$C a C O_{3 (s)} ⇌ C a O_{(s)} + C O_{2 (g)}$ .
Given $K_{p} = 167 ba r$ at $1073 K$ .

$1.896 m o l L^{- 1}$

$4.38 \times 1 0^{- 4} m o l L^{- 1}$

$6.3 \times 1 0^{4} m o l L^{- 1}$

$6.626 m o l L^{- 1}$

$K_{p} = K_{c} (RT)^{Δ n}$ ;
$Δ n = 1$
$K_{p} = 167 ba r$ ,
$K_{c} = \frac{167 ba r}{0.0821 L ba r K ^{- 1} m o l ^{- 1} \times 1073 K}$
$= 1.896 m o l L^{- 1}$

Q. The value of Kc​ for the following equilibrium is CaCO3(s)​⇌CaO(s)​+CO2(g)​. Given Kp​=167bar at 1073K.

1.896molL−1

4.38×10−4molL−1

6.3×104molL−1

6.626molL−1