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Heat of Formation

If 12X2Os→ ...

Question

If $\frac{1}{2} X_{2} O_{(s)} \to X_{(s)} + \frac{1}{4} O_{2 (g)}; Δ H = 90 K J$ ,
then heat change during reaction of metal X with 1 mole $O_{2}$ to form oxide to the maximum extent is

- 360 KJ

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- 180 KJ

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+ 360 KJ

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+ 180 KJ

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Solution

The correct option is D - 360 KJ
$X_{(s)} + \frac{1}{4} O_{2} \to \frac{1}{2} X_{2} O_{(s)}; Δ H = - 90 K J$

To form one more of $O_{2}$ , multiply the above reaction by 4.

The new reaction is:

$4 X_{(s)} + O_{2} \to 2 X_{2} O_{(s)}; Δ H = - 360 K J$

Suggest Corrections

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