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Question

Given, H2S(g)HS(g)+H(g);Hrxn=379.6 kJmol1
H2 (g)+S (s)H2S (g);
HfH2S=21.1 kJmol1
H2(g)2H(g) ; ΔH=438 kJmol1
ΔHsublimation of S = 280 kJmol1

Thus, HfHS is

A
138.5 kJmol1
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B
138.5 kJmol1
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C
277.0 kJmol1
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D
+139.5 kJmol1
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Solution

The correct option is D +139.5 kJmol1
12H2 (g)+S (s)HS (g)
This requires dissociation energy of (HS) bond of HS (g) This can be derived fromfH(H2S).
fH2S (g) = [BE of H2 + sublimation energy of S (s)] - [BE of first HS bond + BE of second HS bond]
21.1=[438+280][379.6+x]x=438+280379.6+21.1=359.5 kJmol1
For the formation of HS (s)
12H2 (g)+S (s)HS (g)
fH(HS)=12BE(H2)+ sublimation energy of S (s) - BE (second HS bond)
=(219+280)(359.5)
=139.5 kJmol1

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