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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

Given the following reaction NH4SH s ltgt NH3 g H2S g If we start with 0085mol NH4SH in a 025L container and are left with 0035 mol of NH4SH at equilibrium calc class=

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SvetkaChem

Answer:

D. 0.3 M

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