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Thermodynamics

Question
CBSEENCH11006222

Calculate the enthalpy change for the process
CCl subscript 4 left parenthesis straight g right parenthesis space rightwards arrow space space straight C left parenthesis straight g right parenthesis space plus space 4 Cl left parenthesis straight g right parenthesis
and calculate the bond enthalpy of C - Cl in CCl4(g).
  space increment subscript vap straight H to the power of 0 left parenthesis CCl subscript 4 right parenthesis space equals space 30.5 space kJ space mol to the power of negative 1 end exponent
space increment subscript straight f straight H to the power of 0 left parenthesis CCl subscript 4 right parenthesis space equals space minus 135.5 space kJ space mol to the power of negative 1 end exponent
space increment subscript straight a straight H to the power of 0 left parenthesis straight C right parenthesis space equals space 715.0 space kJ space mol to the power of negative 1 end exponent comma
where increment subscript straight a straight H to the power of 0 is enthalpy of atomisation and increment subscript straight a straight H to the power of 0 left parenthesis Cl subscript 2 right parenthesis space equals space 242 space kJ space mol to the power of negative 1 end exponent

Solution

We aim at the equation
   CCl subscript 4 left parenthesis straight g right parenthesis space space rightwards arrow space space space straight C left parenthesis straight g right parenthesis space plus space 4 Cl left parenthesis straight g right parenthesis comma
increment straight H equals ?
The given data implies as under
   left parenthesis straight i right parenthesis space CCl subscript 4 left parenthesis straight g right parenthesis space space rightwards arrow space space CCl subscript 4 left parenthesis straight g right parenthesis semicolon space space increment straight H to the power of 0 space equals space 30.5 space kJ space mol to the power of negative 1 end exponent
left parenthesis ii right parenthesis space straight C left parenthesis straight s right parenthesis space plus space 2 Cl subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space CCl subscript 4 left parenthesis straight l right parenthesis semicolon
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment straight H to the power of 0 space equals space minus 135.5 space kJ space mol to the power of negative 1 end exponent
left parenthesis iii right parenthesis space straight C left parenthesis straight s right parenthesis space space rightwards arrow space straight C left parenthesis straight s right parenthesis semicolon space space space increment straight H space equals space 715.0 space kJ space mol to the power of negative 1 end exponent
left parenthesis iv right parenthesis space Cl subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space 2 Cl left parenthesis straight g right parenthesis semicolon space space increment straight H space equals space 242 space kJ space mol to the power of negative 1 end exponent
Multiplying equaton (iv) by 2,
  left parenthesis straight v right parenthesis space 2 Cl subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space space 4 Cl space left parenthesis straight g right parenthesis semicolon space space increment straight H to the power of 0 space equals space 484 space kJ
Adding space equations space left parenthesis iii right parenthesis space and space left parenthesis straight v right parenthesis comma
left parenthesis vi right parenthesis space straight C left parenthesis straight s right parenthesis space plus space 2 Cl subscript 2 left parenthesis straight g right parenthesis space space rightwards arrow space space straight C left parenthesis straight g right parenthesis space plus space 4 Cl left parenthesis straight g right parenthesis semicolon space increment straight H to the power of 0 space equals space 1199 space kJ
Subtracting equation (i) and (ii) from equation (vi),
             straight C left parenthesis straight s right parenthesis space plus space 2 Cl subscript 2 left parenthesis straight g right parenthesis space minus space CCl subscript 4 left parenthesis straight l right parenthesis space minus space straight C left parenthesis straight s right parenthesis space minus space 2 Cl subscript 2 left parenthesis straight g right parenthesis space space
space space space space space space space space space space space space space space space space space space space space space rightwards arrow space straight C left parenthesis straight s right parenthesis space plus space 4 Cl left parenthesis straight g right parenthesis space minus space CCl subscript 4 left parenthesis straight g right parenthesis space minus space CCl subscript 4 left parenthesis straight l right parenthesis
                                 increment straight H space equals space 1199 minus 30.5 space plus space 135.5 space
CCl subscript 4 left parenthesis straight g right parenthesis space rightwards arrow space space space straight C left parenthesis straight g right parenthesis space plus space 4 Cl left parenthesis straight g right parenthesis semicolon space space increment straight H space equals space 1304 space kJ space mol to the power of negative 1 end exponent
Bond space enthalpy space of space straight C space minus space Cl space in space CCl subscript 4 left parenthesis average space value right parenthesis
space space space space space space space space space space space space equals space 1304 over 4 space equals space 326 space kJ space mol to the power of negative 1 end exponent