Question

If the reaction below has a Kp =1.9 × 106 at 400°C, what is Kc?

Answer 1

Answer : The value of $K_c$ is $3.4\times 10^{4}$

Explanation :

The balanced equilibrium reaction is,

$N_2O(g)+NO_2(g)\rightleftharpoons 3NO(g)$

The relation between $K_p$ and $K_c$ are :

$K_p=K_c\times (RT)^{\Delta n}$

where,

$K_p$ = equilibrium constant at constant pressure = $1.9\times 10^6$

$K_c$ = equilibrium concentration constant = ?

R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = $400^oC=273+400=673K$

$\Delta n$ = change in the number of moles of gas = [(3) - (1 + 1)] = 1  (from the chemical reaction)

Now put all the given values in the above relation, we get:

$(1.9\times 10^6)=K_c\times (0.0821L.atm/K.mol\times 673K)^{1}$

$K_c=3.4\times 10^{4}$

Thus, the value of $K_c$ is $3.4\times 10^{4}$