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The equilibrium constant, K, for the following reaction is 1.80x104 at 298 K. NH,HS(s) 2 NH3(g) + H2S(g) Calculate the equilibrium concentration of H,S when 0.237 moles of NH,HS(s) are introduced into a 1.00 L vessel at 298 K. [H,S] = M

The equilibrium constant, K, for the following reaction is 1.80x104 at 298 K. NH,HS(s) 2 NH3(g) + H2S(g) Calculate the equilibrium concentration of H,S when 0.237 moles of NH,HS(s) are introduced into a 1.00 L vessel at 298 K. [H,S] = M

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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The equilibrium constant, K, for the following reaction is 1.80×104 at 298 K. NH,HS(s) 2 NH3(g) + H2S(g) Calculate the equilibrium concentration of H,S when 0.237 moles of NH,HS(s) are introduced into a 1.00 L vessel at 298 K. [H,S] = M
Transcribed Image Text:The equilibrium constant, K, for the following reaction is 1.80×104 at 298 K. NH,HS(s) 2 NH3(g) + H2S(g) Calculate the equilibrium concentration of H,S when 0.237 moles of NH,HS(s) are introduced into a 1.00 L vessel at 298 K. [H,S] = M
Expert Solution
Step 1

The given equilibrium reaction is 

NH_4HS_{(s)} \rightleftharpoons NH_3_{(g)} + H_2S_{(g)}

We can write the expression for the equilibrium constant Kc as follows:

K_c = [NH_3] [H_2S] = 1.80 \times 10^{-4}

Note that NH4HS is in solid state. Hence, its concentration is taken as unity and does not appear in the equilibrium expression.

Volume of the vessel = 1.00 L

Initially, 0.237 mol of NH4HS is taken.

 

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