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Thermodynamics

Question
CBSEENCH11006282

Find out the value of equilibrium constant for the following reaction at 298 K:

2 NH subscript 3 left parenthesis straight g right parenthesis space plus space CO subscript 2 left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space space NH subscript 2 CONH subscript 2 left parenthesis aq right parenthesis space plus space straight H subscript 2 straight O left parenthesis l right parenthesis

   Standard Gibbs energy change,  increment straight G degree at the given temperature, is -13.6 kJ mol-1.

Solution

We know, 
        increment subscript straight r straight G to the power of 0 space equals space minus 2.303 space RT space log space straight K
space space therefore space space space space log space straight K space equals space fraction numerator negative increment subscript straight r straight G to the power of 0 over denominator 2.303 space RT end fraction space space space space space space... left parenthesis 1 right parenthesis
Substituting the values in equation (1), we get
  log space straight K space equals space fraction numerator negative left parenthesis negative 13.6 right parenthesis over denominator 2.303 space left parenthesis 8.314 space cross times space 298 right parenthesis end fraction space equals space 2.38 space
Hence space straight K space equals space antilog space of space 2.38 space equals space 2.4 space cross times space 10 squared