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Find Out the Value of Kc for Each of the Following Equilibria from the Value of Kp: CaCO3 (s) ⇌ CaO(s) + CO2(g); Kp= 167 at 1073 K - Chemistry

Find out the value of Kc for each of the following equilibria from the value of Kp:

CaCO3 (s) ⇌ CaO(s) + CO2(g); Kp= 167 at 1073 K

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Solution 1

`CaCo_3(s) ⇌ CaO(s) + CO_2(g)`

`k_p = 167 atm, triangle^"ng"  =1`

`R = 0.0821 "liter atm" K^(-1) mol^(-1); T = 1073 K`

`K_c = K_p/(RT)^(triangleng) = (167 atm)/(0.0821 L atm K^(-1) mol^(-1) xx 1073 K)^1`

`= 1.9 "mol L"^(-1)`

Solution 2

Here,

Δn = 2 – 1 = 1

R = 0.0831 barLmol–1K–1

T = 1073 K

Kp= 167

Now,

Kp = Kc (RT) Δn

`=> 167 = K_c(0.0831 xx 1073)^(trianglen)`

`=> K_c = 167/(0.0831 xx 1073)`

= 1.87 (approximately)

Concept: Homogeneous Equlibria - Equilibrium Constant in Gaseous Systems
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 7 Equilibrium
Q 5.2 | Page 224
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