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Thermodynamics

Question
CBSEENCH11006279
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The equilibrum constant for a reaction is 10. What will be the value of ∆G0 ? R = 8.314 JK–1 mol–1, T = 300 K. 

Solution

We know, 
        increment straight G to the power of 0 space equals space minus 2.303 space RT space log space straight K Here comma space straight K space equals space 10 semicolon space space straight R equals space 8.314 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent semicolon space space space space space space space space space straight T space space equals space 300 space straight K
Substituting the values, we get,
         increment straight G to the power of 0 space equals space minus 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 space space space space space space space space space space space space space space space space cross times space left parenthesis 300 space straight K right parenthesis space cross times space log space 10 space space space space space equals space minus 5744.1 space straight J space mol to the power of negative 1 end exponent space space

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