Chemistry, asked by laxmanreddysvn, 4 months ago

The value of kc for the following equilibrium is Caco3
Caocs+ Co2cg). Given kp = 167 bar at
1073K

Answers

Answered by s02371joshuaprince47

0

Answer:

1.896

Explanation:

CaCO

3

(s)⇌CaO(s)+CO

2

(g)

Δn=(0+1)−0=1

K

c

=

(RT)

Δn

K

p

K

c

=

(0.0821Lbar/K)×(1073K)

167bar

=1.896mol/L

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