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Calculate the hest of reaction of the following reaction : C_(6)H_(12)O_(6(s)) + rarr 6CO_(2(g)) + 6H_(2)O_((g)) : Delta H = ? |
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Answer» Solution :`"C(graphite)" + O_(2)(g) rarr CO_(2)(g) , Delta H = -395.0 kJ` …(1) `H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l) , Delta H = -269.5 kJ` ...(2) `6"C(graphite)" + 6H_(2)(g) + 3O_(2)(g) rarr C_(6)H_(12)O_(6) , Delta H = -1169.8 kJ` ... (3) Multiplying equation (1) and (2) each by 6 and reversible (3), we get, `6"C(graphite)" + 6O_(2)(g) rarr 6CO_(2(g)) , ""Delta H = -2370 kJ` …(4) `6H_(2(g)) + 3O_(2(g)) rarr 6H_(2)O_((l)) , ""Delta H = -1617.0 kJ`…(5) `C_(6)H_(12)O_(6(s)) rarr 6C_("(graphite)") + 6H_(2(g)) + 3O_(2(g)) , Delta H = +1169.8 kJ` ...(6) ADDING (4), (5) and (6), `C_(6)H_(12)O_(6(s)) + 6O_(2(g)) rarr 6CO_(2(g)) + 6H_(2)O_((g))`, `Delta H(C_(6)H_(12)O_(6)) = -2370.0 - 1617.0 + 1169.8 = -2817.2 kJ`. |
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