The equilibrium constant of a reaction is 8.98e+01 at 442 K and 1.80e+03 at 563

Question

The equilibrium constant of a reaction is 8.98e+01 at 442 K and 1.80e+03 at 563 K. Determine the following for this reaction: Delta H: ____kJ/mol Delta S: ____ J/mol-K What is the value of the equilibrium constant at 503 K? K=_____

Explanation / Answer

Delta G = -RT ln K ......For 442 K ...Delta G = - 8.314 x 442 x ln 89.8 = - 16527.7 J ....for 563 K Delta G = - 35085 J ..... Delta G = Delta H - T x delta S => -16527 = Delta H - 442 x Delta S ....... (1) ........and -35085 = Delta H - 563 x Delta S.....(2) ---------Solving the above two equations we get.....Delta H = 51263.4 J ......and .......Delta S = 153.37 J/K........ At 503 K ....Delta G = -25881.7 J = -RT ln K ....=> K = 493.37 = 4.934e+02

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