1.

For the reaction : C_(2)H_(4)(g) + 3 O_(2)(g) rarr 2 CO_(2)(g) + 2 H_(2)O(l) at 298 K, Delta U = - 1415 kJ. If R = 0.0084 kJ K^(-1). Then Delta H is equal to

Answer»

`- 1400 kJ`
`- 1410 kJ`
`- 1420 kJ`
`- 1430 kJ`.

Solution :`DELTA H = Delta U + Delta n_(G) RT`
`Delta n_(g) = 2 -1 -3 = -2`
`Delta H = - 1415 - 2 xx 0.0084 xx 298`
`= - 1420 kJ`


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