Question

For the reaction
A2(g) + 2B2 ⇌ 2C2(g) the partial pressure of A2, B2 at equilibrium are 0.80 atm and 0.40 atm respectively. The pressure of the system is 2.80 atm. The equilibrium constant Kp will be

(A) 20 (B) 5.0 (C) 0.02 (D) 0.2

Answer 1

Answer:

The correct answer is option A.

Explanation:

$A_2(g) + 2B_2 \rightleftharpoons 2C_2(g)$

Partial pressure of $A_2$ at equilibrium ,$p_1$ = 0.80 atm

Partial pressure of $B_2$ at equilibrium,$p_2$ = 0.40 atm

Partial pressure of $C_2$ at equilibrium,$p_3$ = p

Total pressure at equilibrium = P = 2.80 atm

$P=p_1+p_2+p_3$ (Dalton's law of partial pressures)

$2.80 atm-0.80 atm-0.40 atm=p$

p = 1.60 atm

The expression of $K_p$ will be given as:

$K_p=\frac{p_{3}^2}{p_1\times p_{2}^{2}}=\frac{(1.60 atm)^2}{0.80 atm\times (0.40 atm)^2}=20$

The equilibrium constant Kp will be 20.

Answer 2

Answer : The correct option is, (A) 20

Explanation :

First we have to calculate the partial pressure of $C_2$ gas.

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

$P_T=p_{A_2}+p_{B_2}+p_{C_2}$

where,

$P_T$ = total pressure = 2.80 atm

$p_{A_2}$ = partial pressure of A = 0.80 atm

$p_{B_2}$ = partial pressure of B = 0.40 atm

$p_{C_2}$ = partial pressure of C = ?

Now put all the given values is expression, we get the partial pressure of the $C_2$ gas.

$2.80=0.80+0.40+p_{C_2}$

$p_{C_2}=1.6atm$

Now we have to calculate the value of $K_p$.

The expression of equilibrium constant for the given reaction will be,

$K_p=\frac{(p_{C_2})^2}{(p_{A_2})\times (p_{B_2})^2}$

Now put all the given values in this expression, we get:

$K_p=\frac{(1.6)^2}{(0.80)\times (0.40)^2}$

$K_p=20$

Therefore, the equilibrium constant $K_p$ will be, 20