The value of K p for the reaction, CO 2( g )+ C ( s ) leftharpoons 2 CO ( g ) is 3.0 at 1000 K. If initially P CO 2=0.48 bar and P CO =0 bar and pure graphite is present, calculate the equilibrium partial pressure of CO and CO 2.

Question

The value of K p for the reaction, CO 2( g )+ C ( s ) leftharpoons 2 CO ( g ) is 3.0 at 1000 K. If initially P CO 2=0.48 bar and P CO =0 bar and pure graphite is present, calculate the equilibrium partial pressure of CO and CO 2. (A) 0.15 bar (B)  0.33 bar (C)  0.1475 bar (D)  5.3 bar

Tardigrade Admin · Accepted Answer

The given reaction is :- beginarrayl CO 2( g )+ C ( s ) leftharpoons 2 CO ( g ) ; K P =3 text Initial pressure: 0.48 0 0 text (let) text At eqm: (0.48- p ) text Now, K P =([ CO ]2/[ CO 2]) K P =(( pCO )2/( pCO 2)) ⇒ K P =((2 p )2/(0.48- p )) ⇒ 3(0.48- p )=4 p 2 ⇒ 4 p 2+3 p -1.44=0 ⇒ p =(-3 ± √9+23.04/8)=(-3 ± 5.66/8) ⇒ p =(2.66/8)=0.3325 bar endarray So, Partial pressure of CO at eqm =2 × 0.3325=0.665 bar Partial pressure of CO 2 at eqm =0.48-0.3325=0.1475 bar.

Q. The value of $K_{p}$ for the reaction, $C O_{2} (g) + C (s) ⇌ 2 CO (g)$ is $3.0$ at $1000 K$ . If initially $P_{C O_{2}} = 0.48$ bar and $P_{CO} = 0$ bar and pure graphite is present, calculate the equilibrium partial pressure of $CO$ and $C O_{2}$ .

$0.15$ bar

$0.33$ bar

$0.1475$ bar

$5.3$ bar

Q. The value of Kp​ for the reaction, CO2​(g)+C(s)⇌2CO(g) is 3.0 at 1000K. If initially PCO2​​=0.48 bar and PCO​=0 bar and pure graphite is present, calculate the equilibrium partial pressure of CO and CO2​.

0.15 bar

0.33 bar

0.1475 bar

5.3 bar

Q. The value of $K_{p}$ for the reaction, $C O_{2} (g) + C (s) ⇌ 2 CO (g)$ is $3.0$ at $1000 K$ . If initially $P_{C O_{2}} = 0.48$ bar and $P_{CO} = 0$ bar and pure graphite is present, calculate the equilibrium partial pressure of $CO$ and $C O_{2}$ .

$0.15$ bar

$0.33$ bar

$0.1475$ bar

$5.3$ bar