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

Law of Mass Action

What is minim...

Question

What is minimum mass (in g) of $C a C O_{3} (s)$ will be required to establish the equilibrium $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ at $1000 K$ in a $4 L$ container previously containing $C O_{2}$ at $0.82 a t m$ . Equilibrium constant $K_{P}$ for the dissociation reaction $1.64 a t m$ .

Open in App

Solution

given in this question
T = 1000K
V = 4L
P = 0.82 atm
R 0.083
Neq = $\frac{P V}{R T}$

Neq = $0.82 \times 4 1000 \times 0.08$
= 0.039
$K p = \frac{α^{2}}{X + α}$
$X + α$ = 0.039
so the mole of calcium carbonate is 0.25
so mass is minimum o.25 * 100
= 25 gm

Suggest Corrections

Similar questions

_{3} (s)

6.5

(K_{c} = 0.05 {molL}^{- 1})

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

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

K_{p} = 1.64 a t m

1000 K

50 g

C a C O_{3}

10

1000 K

C a C O_{3}

R = 0.082 L a t m K^{- 1} {m o l}^{- 1}

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

C a C O_{3} (s)

C O_{2}

K_{p}

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

8.21

727^{0}

10

C a C O_{3 (s)}

10

C a C O_{3 (s)}

0.821

1000 K

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

4 \times 10^{- 2} a t m

C a O

Join BYJU'S Learning Program