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Given N(2)(g)+3H(2)(g)to2NH(3)(g),Delta(...

Given `N_(2)(g)+3H_(2)(g)to2NH_(3)(g),Delta_(r)H^(theta)=-92.4 kJ mol^(-1)` What is the standard enthalpy of formation of `NH_(3)` gas ?

A

`-92`

B

`+46`

C

`+92`

D

`-46`

Text Solution

Verified by Experts

The correct Answer is:
D
`N_(2)+3H_2to 2NH_(3)Delta H^(Theta)=-92kJ`.
For molar enthaply formation the chemical reaction of `NH_3` is
`(1)/(2)N_(2)(g)+(3)/(2)H_2(g)to NH_3 " " (because Delta_r H^(Theta) of N_2 and H_2 is 0)`
`Delta _(r) H^(Theta)=(-Delta_r H^Theta)/(2)`
`=(92)/(2)`
`=-46 k j mol^(-1)` .
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