Question

At the start of a reaction, there are 0.0249 mol N2,

Answer 1

Answer:

Explanation:

The reaction is given as:

$N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$

The reaction quotient is:

$Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}$

From the given information:

TO find each entity in the reaction quotient, we have:

$[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}$

$[N_2] = \dfrac{0.024 }{3.5}$

$[N_2] = 0.006857$

$[H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}$

$[H_2] = 9.17 \times 10^{-3}$

∴

$Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135$

However; given that:

$K_c = 1.2$

By relating $Q_c \ \ and \ \ K_c$, we will realize that $Q_c \ \ < \ \ K_c$

The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.