Question

CO(g ) + H2O(g ) <---> CO2(g ) + H2(g ), Kc = 0.0611 at 2000 K . A reaction mixture initially contains a CO partial pressure of 2.50 atm, an H2O partial pressure of 2.50 atm, a CO2 partial pressure of 1.00 atm, and an H2 partial pressure of 1.00 atm at 2000 K. Calculate the equilibrium partial pressure of CO

Answer 1

Answer: Thus equilibrium partial pressure of CO is 1.306 atm

Step-by-step explanation:

`K_p` is the constant of a certain reaction at equilibrium.

For the given chemical reaction:

`CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)`

at t=0 2.50 2.50 1.00 1.00

at em (2.50-x) (2.50-x) (1.00+x) (1.00+x)

The expression of
`K_p` for above equation follows:

`K_p=((p_(CO_2))* p_(H_2))/(p_(CO)* p_(H_2O))`

Putting values in above equation, we get:

`0.0611=((1.00+x)* (1.00+x))/((2.50-x)* (2.50-x))`

`x=0.306`

Thus equilibrium partial pressure of CO = (1.00+x) = (1.00+ 0.306) = 1.306 atm