You visited us 1 times! Enjoying our articles? Unlock Full Access!

Derivation of Kp and Kc

The value of ...

Question

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 $K_{p} = 167$ bar at $1073$ K.

1.896 m o l L^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4.38 \times 10^{- 4} m o l L^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.4 \times 10^{4} m o l L^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.626 m o l L^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1.896 m o l L^{- 1}$
$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$
$Δ n = (0 + 1) - 0 = 1$
$K_{c} = \frac{K_{p}}{(R T)^{Δ n}}$
$K_{c} = \frac{167 b a r}{(0.0821 L b a r / K) \times (1073 K)} = 1.896 m o l / L$

Suggest Corrections

Similar questions

K_{C}

K_{P}

2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g); K_{P} = 1.8 \times 10^{- 2}

500

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g); K_{P} = 167

1073

K_{c}

(i) 2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g); K_{P} = 1.8 \times 10^{- 2} a t 500 K

(i i) C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g); K_{p} = 167 a t 1073 K

R = 0.0831 b a r L m o l^{- 1} K^{- 1}

The equilibrium constant for the reaction,

CaCO₃(s) ⇌ CaO(s) + CO₂(g), is

K_{c}

(i) 2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g); K_{P} = 1.8 \times 10^{- 2} a t 500 K

(i i) C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g); K_{p} = 167 a t 1073 K

R = 0.0831 b a r L m o l^{- 1} K^{- 1}

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g);

Join BYJU'S Learning Program