Question
Formation of ammonia is shown by the reaction, $${N_{2\left( g \right)}} + 3{H_{2\left( g \right)}} \to 2N{H_{3\left( g \right)}};{\Delta _r}{H^ \circ } = - 91.8\,kJ\,mo{l^{ - 1}}$$
What will be the enthalpy of reaction for the decomposition of $$N{H_3}$$ according to the reaction ?
$$2N{H_{3\left( g \right)}} \to {N_{2\left( g \right)}} + 3{H_{2\left( g \right)}};{\Delta _r}{H^ \circ } = ?$$
A.
$$ - 91.8\,kJ\,mo{l^{ - 1}}$$
B.
$$ + 91.8\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 45.9\,kJ\,mo{l^{ - 1}}$$
D.
$$ + 45.9\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ + 91.8\,kJ\,mo{l^{ - 1}}$$
Solution :
The sign of $$\Delta H$$ is reversed.