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For the reaction : C(2)H(4)(g) + 3 O(2...

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

A

`- 1400 kJ`

B

`- 1410 kJ`

C

`- 1420 kJ`

D

`- 1430 kJ`.

Text Solution

Verified by Experts

The correct Answer is:
C
`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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