Question

Given the following reaction: NH4SH (s) <--> NH3 (g) + H2S (g) If we start

with 0.085mol NH4SH in a 0.25L container and are left with 0.035 mol of
NH4SH at equilibrium, calculate k. Options: a)0.02 b) 0.04 c)0.3 and d) 0.3

Answer 1

D. 0.3 M

Step-by-step explanation:

NH4SH (s) <--> NH3 (g) + H2S (g)

Initial concentration 0.085mol/0.25L 0 0

Change in concentration -0.2M +0.2 M +0.2M

Equilibrium 0.035mol/0.25 L=0.14M 0.2M 0.2M

concentration

Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M

K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M