0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

In the reaction : C(s)+CO2(g)2CO(g), the equilibrium pressure is 6 atm. If 50% of CO2 reacts, then Kp of the reaction is _______.

Solution
Verified by Toppr

Let the initial pressure of CO2 be P atm. Since 50% is dissociated, its equilibrium pressure will be 0.5 P atm. The equilibrium pressure of CO will be P atm.

Total pressure will be 0.5P+P=1.5P=6atm

P=4atm

Thus, the equilibrium pressures of CO2 and CO are 2 atm and 4 atm respectively.

The equilibrium constant expression is Kp=P2COPCO2=422=8.

Was this answer helpful?
3
Similar Questions
Q1
In the reaction : C(s)+CO2(g)2CO(g), the equilibrium pressure is 6 atm. If 50% of CO2 reacts, then Kp of the reaction is _______.
View Solution
Q2
In the reaction, C(s)+CO2(g)2CO(g), the equilibrium pressure is 12 atm. If 50% of CO2 reacts, calculate Kp.
View Solution
Q3
In the reaction, C(s)+CO2(g)2CO(g), the equilibrium pressure is 12atm. If 50% of CO2 reacts, calculate Kp
View Solution
Q4
In the reaction C(s)+CO2(g)2CO(g), the equilibrium pressure is 12atm. If 50% of CO2 reacts then Kp will be:
View Solution
Q5
In the reaction C(s)+CO22CO(g), the equilibrium pressure is 12 atm. If 50% CO2 reacts, calculate KP :
View Solution