-->

Equilibrium

Question
CBSEENCH11006415

The value of Kc for the reaction CO left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space space space space CH subscript 3 OH left parenthesis straight g right parenthesis is found to be 186.9 at 750K. Calculate the value of straight K subscript straight p comma if partial pressures are expressed in bars.
        

Solution
The reversible reaction is
                 CO left parenthesis straight g right parenthesis space plus space 2 straight H subscript 2 left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space space space CH subscript 3 OH left parenthesis straight g right parenthesis
We know,     straight K subscript straight P space equals space straight K subscript straight c left parenthesis RT right parenthesis to the power of increment straight n end exponent                   ...(1)
Now,            straight K subscript straight c space equals space 186.9 space
                    straight R space equals space 0.0821 space straight L space bar space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent
straight T space equals space 750 space straight K semicolon space space space space increment straight n space equals space 1 minus left parenthesis 1 plus 2 right parenthesis space equals space minus 2
Substituting the value in equation (1), we have,
    straight K subscript straight p space equals space 186.9 space cross times space left parenthesis 0.0821 space cross times 750 right parenthesis to the power of negative 2 end exponent
space space space space space equals space fraction numerator 186.9 over denominator left parenthesis 0.0821 right parenthesis thin space left parenthesis 750 right parenthesis squared end fraction 0.0494