For the following reaction: 2H 2 S(g) +SO 2 (g) 3S(s)+2H 2 O(g) A reaction mixtu

Question

For the following reaction:    2H2S(g) +SO2(g) 3S(s)+2H2O(g)    A reaction mixture initially containing0.480 M H2S and 0.480 M SO2 was found tocontain 1.1 x 10-3M H2O at a certaintemperature. A Second reaction mixture at the same temperatureinitially contains [H2S] = 0.255 M and[SO2] = 0.335 M.    Calculate the equilibrium concentration ofH2O in the second mixture at this temperature.    2H2S(g) +SO2(g) 3S(s)+2H2O(g)    A reaction mixture initially containing0.480 M H2S and 0.480 M SO2 was found tocontain 1.1 x 10-3M H2O at a certaintemperature. A Second reaction mixture at the same temperatureinitially contains [H2S] = 0.255 M and[SO2] = 0.335 M.    Calculate the equilibrium concentration ofH2O in the second mixture at this temperature.

Explanation / Answer

We know that :      The given Reaction is :                   2H2S(g) + SO2(g)3S(s)+2H2O(g)          intial:  0.48             0.48                 1.13 x 10-3       change:      -2x             -x                       + 2 x      equilibrium : 0.48 -2x         0.48 -x            1.13 x 10-3 + 2 x                    Kc = ( 1.13 x 10-3 + 2 x)2 /   ( 0.48 - 2 x )2 ( 0.48 - x )             2H2S(g) + SO2(g)3S(s)+2H2O(g) intial :  0.255        0.335                    0 change:    - 2x        -x                         + 2 x equilibrium : 0.255 - 2x    0.335 -x           + 2x                                                       Kc =  ( 2 x )2 /   (0.255 - 2 x )2 ( 0.335 - x )          At similartemperature the Kc value remains constant.                  ( 1.13 x 10-3 + 2 x)2 /   ( 0.48 - 2 x )2 ( 0.48 - x )    = ( 2 x )2 /   ( 0.255 - 2 x )2 ( 0.335 - x )                 On solving the above Equation we get :                 X = 0.310062 , 0.386978 M

   The Equilibrium concentration of H2O is 2 x 0.310062 = 0.620124 M                                                               or       2 x 0.193489 = 0.386978 M       change:      -2x             -x                       + 2 x      equilibrium : 0.48 -2x         0.48 -x            1.13 x 10-3 + 2 x                    Kc = ( 1.13 x 10-3 + 2 x)2 /   ( 0.48 - 2 x )2 ( 0.48 - x )             2H2S(g) + SO2(g)3S(s)+2H2O(g) intial :  0.255        0.335                    0 change:    - 2x        -x                         + 2 x equilibrium : 0.255 - 2x    0.335 -x           + 2x                                                       Kc =  ( 2 x )2 /   (0.255 - 2 x )2 ( 0.335 - x )          At similartemperature the Kc value remains constant.                  ( 1.13 x 10-3 + 2 x)2 /   ( 0.48 - 2 x )2 ( 0.48 - x )    = ( 2 x )2 /   ( 0.255 - 2 x )2 ( 0.335 - x )                 On solving the above Equation we get :                 X = 0.310062 , 0.386978 M

   The Equilibrium concentration of H2O is 2 x 0.310062 = 0.620124 M                                                               or       2 x 0.193489 = 0.386978 M

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