Question

Match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	$H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$	(i)	$\Delta H = \Delta U - 2RT$
(B)	$PCl_{5(g)} \rightarrow PCl_{3(g)}\_ Cl_{2(g)}$	(ii)	$\Delta H = \Delta U + 3RT$
(C)	$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$	(iii)	$\Delta H = \Delta U$
(D)	$2N_{2}O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$	(iv)	$\Delta H = \Delta U + RT$

• A. (A)$\rightarrow$ (iii), (B)$\rightarrow$ (i), (C)$\rightarrow$ (ii), (D)$\rightarrow$ (iv)
• B. (A)$\rightarrow$ (iii), (B)$\rightarrow$ (iv), (C)$\rightarrow$ (i), (D)$\rightarrow$ (ii)
• C. (A)$\rightarrow$ (ii), (B)$\rightarrow$ (i), (C)$\rightarrow$ (iv), (D)$\rightarrow$ (iii)
• D. (A)$\rightarrow$ (iv), (B)$\rightarrow$ (ii), (C)$\rightarrow$ (i), (D)$\rightarrow$ (iii)

Answer 1

Answer: (B)

(A)$\rightarrow$ (iii), (B)$\rightarrow$ (iv), (C)$\rightarrow$ (i), (D)$\rightarrow$ (ii)

Answer 2

: (1): $\Delta n_{g} = 2 - 2 = 0$ hence $\Delta H = \Delta U$

(2): $\Delta n_{g} = 2 - 1 = 1$hence $\Delta H = \Delta U + RT$

(3): $\Delta n_{g} = 2 - 4 = - 2$ hence $\Delta H = \Delta U - 2RT$

(4): $\Delta n_{g} = 5 - 2 = 3$ hence $\Delta H = \Delta U + 3RT$