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Thermodynamics

Question
CBSEENCH11006185

What will be the amount of heat evolved when 39 gm of C6H6(l) are burnt? Given that:
straight C subscript 6 straight H subscript 6 left parenthesis straight l right parenthesis space plus space 15 over 2 straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space 3 straight H subscript 2 straight O left parenthesis straight l right parenthesis space plus space 6 CO subscript 2 left parenthesis straight g right parenthesis semicolon
increment straight H space equals space minus 3264.6 space kJ space mol to the power of negative 1 end exponent

Solution
The given thermochemical equation is
  straight C subscript 6 straight H subscript 6 left parenthesis straight l right parenthesis space plus space 15 over 2 straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space 3 straight H subscript 2 straight O left parenthesis straight l right parenthesis space plus space 6 CO subscript 2 left parenthesis straight g right parenthesis semicolon
increment straight H space equals space minus 3264.6 space kJ space mol to the power of negative 1 end exponent
Gram molecular mass of straight C subscript 6 straight H subscript 6
                                 space equals space 6 cross times 12 plus 6 cross times 1 space equals space 78 space straight g
Heat evolved by burning 78g of straight C subscript 6 straight H subscript 6 left parenthesis l right parenthesis
                                       = 3264.6 k J
Heat evolved by burning 39 g straight C subscript 6 straight H subscript 6 left parenthesis straight l right parenthesis space would be = 
fraction numerator 3264.6 space over denominator 78 end fraction cross times 39 space equals space 1632.3 space kJ