Question

The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)Δn where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction:N2(g)+3H2(g)⇌2NH3(g)for which Δn=2−(1+3)=−2.
Part A:For the reaction 2A(g)+2B(g)⇌C(g) Kc = 55.6 at a temperature of 313 ∘C .Calculate the value of Kp.Express your answer numerically.
Part B:For the reaction X(g)+2Y(g)⇌3Z(g)Kp = 3.83×10−2 at a temperature of 119 ∘C .Calculate the value of Kc.

Answer 1

Answer :

(a) The value of $K_p$ is, $5.0\times 10^{-4}$

(b) The value of $K_c$ is, $3.83\times 10^{-2}$

Explanation:

(a) We have to determine the value of $K_p$.

The given balanced reaction is,

$2A(g)+2B(g)\rightleftharpoons C(g)$

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 = ?

$K_c$ = equilibrium concentration constant = 55.6

R = gas constant = 0.08206 L⋅atm/(K⋅mol)

T = temperature = $313^oC=273+313=586K$

$\Delta n$ = change in the number of moles of gas = [1 - (2 + 2)] = -3

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

$K_p=55.6\times (0.08206L.atm/K.mol\times 586K)^{-3}$

$K_p=0.00050=5.0\times 10^{-4}$

The value of $K_p$ is, $5.0\times 10^{-4}$

(b) We have to determine the value of $K_c$.

The given balanced reaction is,

$X(g)+2Y(g)\rightleftharpoons 3Z(g)$

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 = $3.83\times 10^{-2}$

$K_c$ = equilibrium concentration constant = ?

R = gas constant = 0.08206 L⋅atm/(K⋅mol)

T = temperature = $119^oC=273+119=392K$

$\Delta n$ = change in the number of moles of gas = [3 - (2 + 1)] = 0

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

$3.83\times 10^{-2}=K_c\times (0.08206L.atm/K.mol\times 392K)^{0}$

$K_c=3.83\times 10^{-2}$

The value of $K_c$ is, $3.83\times 10^{-2}$