Given, H2S(g)→HS(g)+H(g);△Hrxn=379.6kJmol−1 H2(g)+S(s)→H2S(g); △H∘fH2S=−21.1kJmol−1 H2(g)→2H(g) ; ΔH=438kJmol−1 ΔHsublimation of S = 280kJmol−1
Thus, △H∘fHS is
A
138.5kJmol−1
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B
−138.5kJmol−1
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C
−277.0kJmol−1
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D
+139.5kJmol−1
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Solution
The correct option is D+139.5kJmol−1 12H2(g)+S(s)→HS(g)
This requires dissociation energy of (H−S) bond of HS(g) This can be derived from△fH∘(H2S). △fH2S(g) = [BEofH2 + sublimation energy of S(s)] - [BE of first HS bond + BE of second HS bond] −21.1=[438+280]−[379.6+x]∴x=438+280−379.6+21.1=359.5kJmol−1
For the formation of HS(s) 12H2(g)+S(s)→HS(g) △fH(HS)=12BE(H2)+ sublimation energy of S(s) - BE (second H−S bond) =(219+280)−(359.5) =139.5kJmol−1