The equilibrium partial pressures of CO(g), `CO_(2)(g)` in the equilibrium reaction
`CO_(2)(g)+C(s) iff 2CO(g)` at 1000 K are 0.66 and 0.15 bar respectively. The equilibrium constant `K_(c)` are apporximately is

Options

A. 0.35
B. 2.9
C. 0.035
D. 0.29

Answer

C

Explanation

Given,
`" "CO_(2)(g)+C(s) iff 2CO(g)`
Equilibrium partial pressure of `CO_(2)(p_(CO_(2)))=0.15` bar
Equilibrium partial of `CO(p_(CO))=0.66` bar
Temperature (T) = 1000 K
For this basic equation, all of the products and reactants are gases.
The equation will be
`" "K_(p)=(p_(CO))^(2)/(P_(CO_(2)))=(0.66)^(2)/0.15=2.904`
`K_(p)` can also be obtained from `K_(C)`.
`K_(C)` is the concentration equilibrium constant, as shown
`" "K_(p)=K_(C)(RT)^(Deltan)`
Here, `" "Deltan=1`
`" "T=1000K`
`" "R=0.0821" atm L "/" mol K"`
Put the value in equation (i)
`" "2904=K_(C)(0.082 times 1000)^(1)`
`" "K_(C)=2.904/(0.082 times 1000)=0.0353`