Hydrogen sulfide decomposes according to the followingreaction, for which K c =

Question

Hydrogen sulfide decomposes according to the followingreaction, for which Kc = 9.30 x 10-8 at 700degrees Celsius. 2 H2S (g) --> <-- 2 H2 (g) +S2 (g) if 0.47 mol of H2S is placed in a 3.0 L container,what is the equilibrium concentration of H2 (g) at 700degress Celsius? please help me solve this problem, including the solution. Ikeep getting it wrong. Thanks in advance :D Hydrogen sulfide decomposes according to the followingreaction, for which Kc = 9.30 x 10-8 at 700degrees Celsius. 2 H2S (g) --> <-- 2 H2 (g) +S2 (g) if 0.47 mol of H2S is placed in a 3.0 L container,what is the equilibrium concentration of H2 (g) at 700degress Celsius? please help me solve this problem, including the solution. Ikeep getting it wrong. Thanks in advance :D

Explanation / Answer

concentration of H2S = Moles / volume                                  =0.47 / 3.0 L                                  =0.156 M                   2H2S (g) <----> 2 H2 (g) +S2 (g) Initial            0.156                      0            0 Change            -2x                     +2x         +x Equilibrium   (0.156 -2x)            2x            x Kc = 9.30 x 10-8 = 2x. x / (0.156 - 2x) Solving, we get x = 8.5 * 10 -5 So equilibrium concentration of H2 = 2x = 2 (  8.5 *10 -5)                                                              =1.7 * 10-4 M So equilibrium concentration of H2 = 2x = 2 (  8.5 *10 -5)                                                              =1.7 * 10-4 M

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