Home
Class 12
CHEMISTRY
The value of K(p) for the reaction. CO...

The value of `K_(p)` for the reaction.
`CO_(2)(g) + C(s) hArr 2CO(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 pressures of `CO` and `CO_(2)`.

A

`P_(CO)=0.15bar, P_(CO_(2))= 0.66 bar`

B

`P_(CO)=0.66 bar, P_(CO_(2))= 0.15 bar`

C

`P_(CO)=0.33 bar, P_(CO_(2))= 0.66 bar`

D

`P_(CO)= 0.66 bar, P_(CO_(2)) = 0.33 bar`

Text Solution

Verified by Experts

The correct Answer is:
B
The value of `K_(p)` for the reaction …………..
`{:(,CO_(2)(g)+,C(s)hArr2,CO(g)),(t=0,0.48,-,0),(t=equi^(m),0.48-x,-,2x):}`
`K_(p) = ((2x)^(2))/((0.48-x)) = 3`
`x = 0.33 bar`
Promotional Banner

Similar Questions

Explore conceptually related problems

The value of K_(p) for the reaction CO_(2)(g)+C(s)hArr2CO(g) is 3.0 bar at 1000 K . If initially P_(CO_(2)) = 0.48 bar, P_(CO) = 0 bar and pure graphite is present then determine equilibrium partial pressue of CO and CO_(2) .

K_(p)//K_(c) for the reaction CO(g)+1/2 O_(2)(g) hArr CO_(2)(g) is

K_(p)//K_(c) for the reaction CO(g)+1/2 O_(2)(g) hArr CO_(2)(g) is

At 1,000 K in the reaction CO_(2) (g) + C(s) rarr 2CO(g) The value of P_(CO_(2)) =0.48" bar and" P_(CO) = 0 bar. Pure graphite is present. The equilibrium partial pressures of CO and CO_(2) are 0.66 bar and 0.15 bar respectively. Calculate K_(P) of the reaction.

For the reaction CaCO_(3(s)) hArr CaO_((s))+CO_(2(g))k_p is equal to

For the reaction , CO(g)+Cl(g)hArrCOCl_2(g) then K_p//K_c is equal to :

For the reaction, CO(g) +(1)/(2) O_2(g) hArr CO_2 (g), K_p//K_c is equal to

For the reaction CO(g)+(1)/(2) O_(2)(g) hArr CO_(2)(g),K_(p)//K_(c) is

For the reactions, CO(g) +Cl_2( g) hArr COCl_2(g), " the " (K_P)/(K_c) is equal to

The equilibrium constant K_(p) for the reaction H_(2)(g)+CO_(2)(g)hArrH_(2)O(g)+CO(g) is 4.0 at 1660^(@)C Initially 0.80 mole H_(2) and 0.80 mole CO_(2) are injected into a 5.0 litre flask. What is the equilibrium concentration of CO_(2)(g) ?