Question

Consider the following reaction where Kc = 0.159 at 723 K. N2(g) + 3H2(g) 2NH3(g) A reaction mixture was found to contain 1.97×10-2 moles of N2(g), 3.82×10-2 moles of H2(g) and 5.27×10-4 moles of NH3(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc, equals 9.63x10^4 . The reaction b A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.

Answer 1

Answer:

The reaction moves to the left. To the reactants side

Explanation:

The reaction given by the problem is:

$N_2_(_g_)~+~3H_2_(_g_)~<=>~2NH_3_(_g_)$

Therefore the equilibrium constant expression would be:

$K_e_q=\frac{[NH_3]^2}{[N_2][H_2]^3}$

With the values equilibrium concentration values we can calculate the quotient "Qc", so:

$Q_c=\frac{[5.27X10^-^4]^2}{[1.97X10^-^2][3.82X10^-^2]^3}=0.251$

In this case Qc>Kc we will have more amount of product. Therefore, the reaction will go to the left to reach the equilibrium.

I hope it helps!