-->

Thermodynamics

Question
CBSEENCH11006265

For the reaction
2 NO left parenthesis straight g right parenthesis space plus space straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards arrow space space space 2 NO subscript 2 left parenthesis straight g right parenthesis comma
calculate increment straight G at 700K when enthalpy and entropy changes are -113 kJ mol-1 and -145 JK-1 mol-1 respectively.

Solution

  Here comma space space increment straight H space equals space minus 113 space kJ space mol to the power of negative 1 end exponent
space space space space space space space space space space space space space space space space equals negative 11300 space straight J space mol to the power of negative 1 end exponent
space space space space space space space space space space space increment straight S space equals space minus 145 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent
According to Gibb's Helmoltz equation,  
 space space space space space space space space increment straight G space equals space increment straight H minus straight T increment straight S
therefore space space space space space space space increment straight G space equals space minus 113000 space 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 space space space equals space minus left parenthesis 700 straight K right parenthesis space cross times space left parenthesis negative 145 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 equals space minus 113000 space straight J space mol to the power of negative 1 end exponent space plus space 101500 space 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 space space space equals space minus 11500 space straight J space mol to the power of negative 1 end exponent