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InterviewSolution
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Calculate the standard heat of formation of carbon disulphide (l). Given that the standard heats of combustion of carbon (s), sulphur (s) and carbon disulphide (l) are -393.3, -293.72 and -1108.76 kJ `mol^(-1)` respectively.A. `-128.02 "KJ mole"^(-1)`B. `+12.802 "KJ mol"^(-1)`C. `+128.02 "KJ mol"^(-1)`D. `-12.802 KJ mol"^(-1)` |
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Answer» Correct Answer - C `C_((s))+2S_((s))rarr CS_(2(l)) , Delta H_(f) = ?` `Delta H_(f)=Sigma (Delta H_("Comb."))_(R )-Sigma (Delta H_("Comb."))_(P)` `Delta H_(f)=[1xx(-393.3)+2xx(-293.72)]-(-1108.76)` `Delta H_(f)=+128.02 kJ mol^(-1)` |
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