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InterviewSolution
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The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are `- 890 .3 kJmol^(-1), -393.5kJ mol^(-1)` and `- 28.5 kJ mol^(-1)` respectively. Enthalpy of formation of `CH_(4)(g)` will be `:` (i) `-74.8 kJ mol^(-1)` (ii) `-52.27 kJ mol^(-1)` (iii) `+74.8 kJ mol^(-1)` (iv) `+ 52 .26 kJ mol^(-1)` |
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Answer» Given `: ` (i) `CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l), DeltaH = - 890 . 3 kJ mol^(-1)` (ii) ` C(s) + O_(2)(g)rarr CO_(2)(g) , DeltaH= -393.5 kJ mol^(-1)` (iii) `H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(l), DeltaH = - 285.8 kJ mol^(-1)` Aim `: C(s) + 2H_(2)(g) rarr CH_(4)(g) , DeltaH = ?` Eqn. (ii) `+ 2 xx` Eqn. (iii) - Eqn. (i) gives the required equation with `DeltaH = - 393.5 +2 ( - 285.8) - ( -890.3 ) kJ mol^(-1)` `= - 74.8 kJ mol^(-1)` Hence, (i) is the correct answer. |
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