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Q.

Given thatOg+e-→ O-g;        ∆H=34 kcal mol-1 Og+2e-→ O2-g;        ∆H=168 kcal mol-1 The enthalpy change for the reactionO-g+e→ O2-g is

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a

-202 kcal mol-1

b

-134 kcal mol-1

c

+134 kcal mol-1

d

+202 kcal mol-1

answer is C.

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Detailed Solution

Given,Og+e-→ O-g;        ∆H=34 kcal mol-1 ∴  O-g+→ Og+ e-;        H=-34 kcal mol-1    ...... i Og+2e-→ O2-g;        ∆H=168 kcal mol-1    .....ii i+ii Og-+e-→O2-g;   ∆H=-34+168 =134 k cal mol-1
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