20 g of sample of CaCO 3( s ) of 50 % impurity is placed in a sealed vessel of 1 L capacity at 400 K when 50 % T dissociation takes place. If CO 2 gas at 0.5 atm is already present, percent dissociation of CaCO 3 is CaCO 3( s ) leftharpoons CaO ( s )+ CO 2( g )

Question

20 g of sample of CaCO 3( s ) of 50 % impurity is placed in a sealed vessel of 1 L capacity at 400 K when 50 % T dissociation takes place. If CO 2 gas at 0.5 atm is already present, percent dissociation of CaCO 3 is CaCO 3( s ) leftharpoons CaO ( s )+ CO 2( g ) (A) 25.0 % (B) 28.7 % (C) 40.0 % (D) 34.8 %

Tardigrade Admin · Accepted Answer

x=0.05 mol (50 % of 0.1 mol ) pC O2=(n/V) RT=(0.05/1) × 0.0821 × 400 =1.642 atm Thus, K p =pC O2=1.642

Kp=(0.5+x)=1.642 ∴ x=1.142 p=(n R T/V) 1.142 atm=(n × 0.0821 × 400/1) n =0.0348 (dissociated) Thus, percentage =(0.0348/0.1) × 100=34.8 %

Q. $20 g$ of sample of $C a C O_{3} (s)$ of $50%$ impurity is placed in a sealed vessel of $1 L$ capacity at $400 K$ when $50% T$ dissociation takes place. If $C O_{2}$ gas at $0.5 a t m$ is already present, percent dissociation of $C a C O_{3}$ is
$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$

$25.0%$

$28.7%$

$40.0%$

$34.8%$

Q. 20g of sample of CaCO3​(s) of 50% impurity is placed in a sealed vessel of 1L capacity at 400K when 50%T dissociation takes place. If CO2​ gas at 0.5atm is already present, percent dissociation of CaCO3​ is CaCO3​(s)⇌CaO(s)+CO2​(g)

25.0%

28.7%

40.0%

34.8%

x=0.05mol(50% of 0.1mol) pCO2​​=Vn​RT=10.05​×0.0821×400 =1.642atm Thus, Kp​=pCO2​​=1.642 Kp​=(0.5+x)=1.642 ∴x=1.142 p=VnRT​ 1.142atm=1n×0.0821×400​ n=0.0348 (dissociated) Thus, percentage =0.10.0348​×100=34.8%

Q. $20 g$ of sample of $C a C O_{3} (s)$ of $50%$ impurity is placed in a sealed vessel of $1 L$ capacity at $400 K$ when $50% T$ dissociation takes place. If $C O_{2}$ gas at $0.5 a t m$ is already present, percent dissociation of $C a C O_{3}$ is
$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$

$25.0%$

$28.7%$

$40.0%$

$34.8%$