ThermochemistryMedium
Question
The data below refers to gas phase reaction at constant pressure at 25°C.
CH3–CH3 → CH3–CH2 + H: ΔH1 = +420 kJ mol–1
CH3–CH2 → CH2 = CH2 + H: ΔH2 = +168 kJ mol–1
From these data, the enthalpy change ΔH for the reaction
2CH3–CH2 → CH3–CH3 + CH2 = CH2 is
Options
A.+250 kJ
B.+588 kJ
C.−252 kJ
D.−588 kJ
Solution
Given T.C.E. are
$(a)CH_{3} - CH_{3} \rightarrow CH_{3} - CH_{3} + H;\Delta H = 420\text{ kJ} $$$(b)CH_{3} - CH_{2} \rightarrow CH_{2} = CH_{2} + H;\Delta H = 168\text{ kJ}$$
Required T.C.E. are
$2CH_{3} - CH_{2} \rightarrow CH_{3} - CH_{3} + CH_{2} = CH_{2};\Delta H = ? $$$\text{From }(b) - (a):\Delta H = 168 - 420 = - 252\text{ kJ}$$
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