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Thermodynamics

Question
CBSEENCH11006268

Calculate the free energy change of the reaction straight C left parenthesis graphite right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space space CO subscript 2 left parenthesis straight g right parenthesis
left parenthesis Given space increment straight H space equals space minus 300 space kJ space mol to the power of negative 1 end exponent semicolon space space increment straight S space equals space 3 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent space at space 300 space straight K right parenthesis.

Solution

According to Gibb's Helmoltz equation,
                  increment straight G space equals space increment straight H space minus space straight T increment straight S space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Here          increment straight H space equals space minus 300 space kJ space mol to the power of negative 1 end exponent
                        equals space minus 30000 space straight J space mol to the power of negative 1 end exponent
                        increment straight S space equals space 3 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent space space and space straight T space equals space 300 space straight K
Substituting the values in (1), we have,
   increment straight G space equals space minus 30000 space straight J space mol to the power of negative 1 end exponent space minus space left parenthesis 300 space straight K right parenthesis space cross times space left parenthesis 3 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 equals space minus 30000 space straight J space mol to the power of negative 1 end exponent space minus 900 space straight J space mol to the power of negative 1 end exponent
space space space space space space space equals space minus 30900 space straight J space mol to the power of negative 1 end exponent
or space space space space minus 30.9 space kJ