For the reaction, 1 g mole of CaCO 3 is enclosed in 5 L container CaCO 3(s) longrightarrow CaO (s)+ CO 2(g) Kp=1.16 at 1073 K then per cent dissociation of CaCO 3 is

Question

For the reaction, 1 g mole of CaCO 3 is enclosed in 5 L container CaCO 3(s) longrightarrow CaO (s)+ CO 2(g) Kp=1.16 at 1073 K then per cent dissociation of CaCO 3 is (A) zero (B) 6.58% (C) 65% (D) 100%

Tardigrade Admin · Accepted Answer

CaCO 3( s ) longrightarrow CaO ( s )+ CO 2(g) ks= PCO3 (only gaseous molecule count) Kp=Kc(R T)Δ n 1.16 =(x/5)(0.0821 × 1073)1 x =(1.16 × 5/0.0821 × 1073)=0.0658 =0.0658 × 100=6.58 %

Q. For the reaction, $1 g$ mole of $C a C O_{3}$ is enclosed in $5 L$ container
$C a C O_{3} (s) ⟶ C a O (s) + C O_{2} (g)$
$K_{p} = 1.16$ at $1073 K$ then per cent dissociation of $C a C O_{3}$ is

zero

6.58%

65%

100%

$C a C O_{3} (s) ⟶ C a O (s) + C O_{2} (g)$
$k_{s} = P_{C O_{3}}$ (only gaseous molecule count)
$K_{p} = K_{c} (RT)^{Δ n}$
$1.16 = \frac{x}{5} (0.0821 \times 1073)^{1}$
$x = \frac{1.16 \times 5}{0.0821 \times 1073} = 0.0658$
$= 0.0658 \times 100 = 6.58%$

Q. For the reaction, 1g mole of CaCO3​ is enclosed in 5L container CaCO3​(s)⟶CaO(s)+CO2​(g) Kp​=1.16 at 1073K then per cent dissociation of CaCO3​ is

zero

6.58%

65%

100%

CaCO3​(s)⟶CaO(s)+CO2​(g) ks​=PCO3​​ (only gaseous molecule count) Kp​=Kc​(RT)Δn 1.16=5x​(0.0821×1073)1 x=0.0821×10731.16×5​=0.0658 =0.0658×100=6.58%

Q. For the reaction, $1 g$ mole of $C a C O_{3}$ is enclosed in $5 L$ container
$C a C O_{3} (s) ⟶ C a O (s) + C O_{2} (g)$
$K_{p} = 1.16$ at $1073 K$ then per cent dissociation of $C a C O_{3}$ is

$C a C O_{3} (s) ⟶ C a O (s) + C O_{2} (g)$
$k_{s} = P_{C O_{3}}$ (only gaseous molecule count)
$K_{p} = K_{c} (RT)^{Δ n}$
$1.16 = \frac{x}{5} (0.0821 \times 1073)^{1}$
$x = \frac{1.16 \times 5}{0.0821 \times 1073} = 0.0658$
$= 0.0658 \times 100 = 6.58%$