Question
Match the column I with column II and mark the appropriate choice.
|
Column I |
|
Column II |
| a. |
$${H_{2\left( g \right)}} + B{r_{2\left( g \right)}} \to 2HB{r_{\left( g \right)}}$$ |
1. |
$$\Delta H = \Delta U - 2RT$$ |
| b. |
$$PC{l_{5\left( g \right)}} \to PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}}$$ |
2. |
$$\Delta H = \Delta U + 3RT$$ |
| c. |
$${N_{2\left( g \right)}} + 3{H_{2\left( g \right)}} \to 2N{H_{3\left( g \right)}}$$ |
3. |
$$\Delta H = \Delta U$$ |
| d. |
$$2{N_2}{O_{5\left( g \right)}} \to 4N{O_{2\left( g \right)}} + {O_{2\left( g \right)}}$$ |
4. |
$$\Delta H = \Delta U + RT$$ |
A.
a - 3, b - 1, c - 2, d - 4
B.
a - 3, b - 4, c - 1, d - 2
C.
a - 2, b - 1, c - 4, d - 3
D.
a - 4, b - 2, c - 1, d - 3
Answer :
a - 3, b - 4, c - 1, d - 2
Solution :
$$\left( {\text{A}} \right):\Delta {n_g} = 2 - 2 = 0\,;$$ hence $$\Delta H = \Delta U$$
$$\left( {\text{B}} \right):\Delta {n_g} = 2 - 1 = 1\,;$$ hence $$\Delta H = \Delta U + RT$$
$$\left( {\text{C}} \right):\Delta {n_g} = 2 - 4 = - 2\,;$$ hence $$\Delta H = \Delta U - 2RT$$
$$\left( {\text{D}} \right):\Delta {n_g} = 5 - 2 = 3;$$ hence $$\Delta H = \Delta U + 3RT$$