Question

For the reaction A(g)+B(g) equilibrium with C(g)+D(g) and kc = 50 Mol per litre at 127 degree celsius then calculate kp?

Answer 1

Here we need to use the relation b/w Kp and Kc

Answer 2

Explanation:

There is following relationship between K_{p}K

p

and K_{c}K

c

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

p

=49∗0.0821∗(127+273)=1.61∗10

3

Conclusion:

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

p

=1.61∗10

3