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Thermodynamics

Question
CBSEENCH11006270

At what temperature, reduction of lead oxide to lead by carbon
                                PbO left parenthesis straight s right parenthesis space plus space straight C left parenthesis straight s right parenthesis space space space rightwards arrow space space Pb left parenthesis straight s right parenthesis space plus space CO left parenthesis straight g right parenthesis
becomes spontaneous? For this reason increment straight H space and space increment straight S are 108.4 kJ mol-1 and 190.0 JK-1 mol-1 respectively. increment straight G space equals space increment straight H space minus space straight T increment straight S

Solution

According to Gibb's Helmoltz equation  increment straight G space equals space increment straight H space minus space straight T increment straight S
At equilibrium,
 increment straight G space equals space 0
space space space space space 0 space equals space increment straight H space minus straight T increment straight S
or space space straight T space equals space fraction numerator increment straight H over denominator increment straight S end fraction                   
  Here space increment straight H space equals space 108.4 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 equals space 108400 space straight J space mol to the power of negative 1 end exponent
space space increment straight S space equals space 190.0 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent
therefore space space space space straight T space equals space fraction numerator 108400 space straight J space mol to the power of negative 1 end exponent over denominator 190.0 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent end fraction space equals space 570.5 space straight K
Above this temperature, ∆G will be negative and the reaction will become spontaneous.