Question

The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.20 atm and H2 is 0.15 atm?

Answer 1

Answer:

"$6.7\times 10^{-4} \ atm$" is the right answer.

Explanation:

Given:

Partial pressure of $N_2$,

= 0.20 atm

Partial pressure of $H_2$,

= 0.15 atm

$K_p = 1.5\times 10^3$ at $400^{\circ} C$

As we know,

⇒ $K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}$

By putting the values, we get

    $1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}$

        $pNH_3^2 = \frac{0.000675}{1.5\times 10^3}$

                    $=6.7\times 10^{-4} \ atm$