1.

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 = ?

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`.


Discussion

No Comment Found