Convert 1 mole to 1 gramif the value of deltah combustion (c6h6)
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Information shadow: (i) C6H6(l)+152O2(g)→6CO2(g)+3H2O(l); ΔH=−3267kJ (ii) C(s)+O2(g)→CO2(g);ΔH=−393.5kJ (iii) H2(g)+12O2(l)→H2O(l);ΔH=−285.83kJ The required equation is: 6C(s)+3H2(g)→C6H6(l);ΔH=? Problem solving strategy: The heat of required equation can be obtained by algebraic method. (ii) × 6 + (iii) × 3 + (i) Working it out: 6C(s)+6O2(g)→6CO2(g);ΔH=−393.5×6kJ3H2(g)+32O2(g)→3H2(l);ΔH=−285.83×3kJ6CO2(g)+3H2O(l)→C6H6(l)+152O2(g); ΔH=+3267kJ On adding,_____________________________ 6C(s)+3H2(g)→C6H6(l);ΔH=+48.51kJmol−1Alternatively, 6C(s)+3H2(g)→C6H6(l) ΔrH=∑Heat of combustion of reactants −∑ Heat of combustion of products = 6 x (-393.5) + 3 (-285.83) - (- 3267) =48.51kJmol−1
Information shadow: (i) C6H6(l)+152O2(g)→6CO2(g)+3H2O(l); ΔH=−3267kJ (ii) C(s)+O2(g)→CO2(g);ΔH=−393.5kJ (iii) H2(g)+12O2(l)→H2O(l);ΔH=−285.83kJ The required equation is: 6C(s)+3H2(g)→C6H6(l);ΔH=? Problem solving strategy: The heat of required equation can be obtained by algebraic method. (ii) × 6 + (iii) × 3 + (i) Working it out: 6C(s)+6O2(g)→6CO2(g);ΔH=−393.5×6kJ3H2(g)+32O2(g)→3H2(l);ΔH=−285.83×3kJ6CO2(g)+3H2O(l)→C6H6(l)+152O2(g); ΔH=+3267kJ On adding,_____________________________ 6C(s)+3H2(g)→C6H6(l);ΔH=+48.51kJmol−1Alternatively, 6C(s)+3H2(g)→C6H6(l) ΔrH=∑Heat of combustion of reactants −∑ Heat of combustion of products = 6 x (-393.5) + 3 (-285.83) - (- 3267) =48.51kJmol−1
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