Question

For a reaction, A(g)+B(s) equilibrium C(g)+D(g) kc=49 mol per litre at 127 degree Celsius ...calculate Kp

Answer 1

Concept :

There is following relationship between $K_{p}$ and $K_{c}$

$K_{p} = K_{c}*(RT)^{n}$

Here,

n = [total number of moles of product - total number of moles of reactant ]

Note: we consider moles only gaseous species not for solid or liquids

Explanation:

According to given equilibrium reaction,

n = 2-1 = 1

On substituting given values in above formula we get ,

$K_{p} = 49*0.0821*(127+273)= 1.61*10^{3}$

Conclusion:

$K_{p} = 1.61*10^{3}$