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Question

If $$H_{2}S+\frac{3}{2}O_{2}\rightarrow H_{2}O+SO_{2},$$
$$\Delta H=-136.7Kcal$$
$$H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O,\Delta H=-68.4Kcal$$
$$S+O_{2}\rightarrow SO_{2},\Delta H=-71Kcal$$
what is $$\Delta H_{f} of H_{2}S$$


A
2.7 Kcal
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B
-2.7 Kcal
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C
5.4 Kcal
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D
-5.4 Kcal
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Solution

The correct option is B -2.7 Kcal
(i) $$H_{2}S+\frac{3}{2}O_{2}\rightarrow H_{2}O+SO_{2}-136.7$$ Kcal
(ii) $$H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2}O-68.4$$ Kcal
(iii) $$S+O_{2}\rightarrow SO_{2}-71$$ Kcal
(ii) + (iii) - (i)
$$H_{2}+S\rightarrow H_{2}S$$
$$\Delta H=$$ -68.4 - 71 + 136.7
$$=$$ - 2.7 Kcal

Chemistry

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