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Thermodynamics

Question
CBSEENCH11006285

Using the data given below, calculate the value of equilibrium constant for the reaction

stack 3 HC identical to CH left parenthesis straight g right parenthesis with Acetylene below space space rightwards harpoon over leftwards harpoon space space straight C subscript 6 straight H subscript 6 left parenthesis straight g right parenthesis space at space 298 space straight K

assuming ideal behaviour.

increment subscript straight r straight G to the power of 0 space left parenthesis HC space identical to space CH right parenthesis space equals space 2.09 space cross times space 10 to the power of 5 space straight J space mol to the power of negative 1 end exponent comma space
increment subscript straight r straight G to the power of 0 left parenthesis straight C subscript 6 straight H subscript 6 right parenthesis space equals space 1.24 space cross times space 10 to the power of 5 space straight J space mol to the power of negative 1 end exponent space and
straight R space equals space 8.314 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent                       

Solution

We know, 
     increment subscript straight r straight G to the power of 0 space equals space sum from blank to blank of increment subscript straight f straight G to the power of 0 left parenthesis products right parenthesis space minus space sum from blank to blank of increment subscript straight f straight G to the power of 0 left parenthesis reactants right parenthesis
equals space open square brackets increment straight G to the power of 0 left parenthesis straight C subscript 6 straight H subscript 6 right parenthesis close square brackets space minus space open square brackets 3 space cross times space increment subscript straight f straight G to the power of 0 left parenthesis HC space identical to CH right parenthesis close square brackets
equals space 1.24 space cross times space 10 to the power of 5 straight J space mol to the power of negative 1 end exponent space minus space 3 space cross times space left parenthesis 2.09 space cross times space 10 to the power of 5 space straight J space mol to the power of negative 1 end exponent right parenthesis
space equals space minus 5.03 space cross times space 10 to the power of 5 space straight J space mol to the power of negative 1 end exponent
  Now comma space increment subscript straight r straight G to the power of 0 space equals space minus 2.303 space RT space log space straight K
or space space space space space minus 5.03 space cross times space 10 to the power of 5 space straight J space mol to the power of negative 1 end exponent
equals negative 2.303 space cross times space left parenthesis 8.314 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent right parenthesis space cross times space left parenthesis 298 space straight K right parenthesis space log space straight K
therefore space space log space straight K space equals space fraction numerator 5.03 space cross times space 10 to the power of 5 straight J space mol to the power of negative 1 end exponent over denominator left parenthesis 2.303 space cross times space 8.314 space cross times space 298 right parenthesis end fraction straight J space mol to the power of negative 1 end exponent
space space space space space space space space space space space space space equals space 88.15
therefore space space space space straight K space equals space Antilog space 88.155 space equals space 1.429 space cross times space 10 to the power of 88