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Question
The enthalpy of combustion of benzene from the following data will be:
(i) 6C(s)+3H2(g)→C6H6(l);ΔH=+45.9kJ
(ii) H2(g)+12O2(g)→H2O(l);ΔH=−285.9kJ
(iii) C(s)+O2(g)→CO2(g);ΔH=−393.5kJ
The enthalpy of combustion of benzene from the following data will be:
(i) 6C(s)+3H2(g)→C6H6(l);ΔH=+45.9kJ
(ii) H2(g)+12O2(g)→H2O(l);ΔH=−285.9kJ
(iii) C(s)+O2(g)→CO2(g);ΔH=−393.5kJ
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Solution
The correct option is D −3264.6 kJ
The combustion of benzene is represented by equation (iv)
(i) 6C(s)+3H2(g)→C6H6(l);ΔH=+45.9kJ
(ii) H2(g)+12O2(g)→H2O(l);ΔH=−285.9kJ
(iii) C(s)+O2(g)→CO2(g);ΔH=−393.5kJ(iv) C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)Reverse equation (i)(v) C6H6(l)→6C(s)+3H2(g);ΔH=−45.9kJMultiply equation (ii) with 3.(vi) 3H2(g)+32O2(g)→3H2O(l);ΔH=−285.9×3=−857.7kJMultiply equation (iii) with 6.(vii) 6C(s)+6O2(g)→6CO2(g);ΔH=−393.5×6=−2361kJAdd equations (v), (vi) and (vii)(iv) C6H6(l)+152O2(g)→CO2(g)+3H2O(l);ΔH=−45.9−857.7−2361=−3264.6kJThe enthalpy of combustion of benzene is −3264.6kJ.
The combustion of benzene is represented by equation (iv)
(i) 6C(s)+3H2(g)→C6H6(l);ΔH=+45.9kJ
(ii) H2(g)+12O2(g)→H2O(l);ΔH=−285.9kJ
(iii) C(s)+O2(g)→CO2(g);ΔH=−393.5kJ
(iv) C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)
Reverse equation (i)
(v) C6H6(l)→6C(s)+3H2(g);ΔH=−45.9kJ
Multiply equation (ii) with 3.
(vi) 3H2(g)+32O2(g)→3H2O(l);ΔH=−285.9×3=−857.7kJ
Multiply equation (iii) with 6.
(vii) 6C(s)+6O2(g)→6CO2(g);ΔH=−393.5×6=−2361kJ
Add equations (v), (vi) and (vii)
(iv) C6H6(l)+152O2(g)→CO2(g)+3H2O(l);ΔH=−45.9−857.7−2361=−3264.6kJ
The enthalpy of combustion of benzene is −3264.6kJ.
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