Question

# The value of $$K_c$$ for the following equilibrium is $$CaCO_{3(s)} \rightleftharpoons CaO_(s) + CO_{2(g)}$$.Given $$K_p =167$$ bar at $$1073$$K.

A
1.896molL1
B
4.38×104molL1
C
6.4×104molL1
D
6.626molL1

Solution

## The correct option is A $$1.896 mol L^{-1}$$$$CaCO_3(s) \rightleftharpoons CaO(s) +CO_2(g)$$$$\Delta n = (0+1)-0 =1$$$$K_c = \dfrac {K_p}{(RT)^{\Delta n}}$$$$K_c = \dfrac {167bar} {(0.0821Lbar/K) \times (1073K)}= 1.896 mol/L$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More