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The heat change for the following reaction at 298 K and constant pressure is +7.3 kcal. $A_2B(s) \rightarrow 2A(s) + \frac{1}{2} B_2(g)$; $\Delta H = + 7.3 kcal$ The heat change at constant volume would be

(a) +7.3 kcal
(b) More than +7.3 kcal
(c) Less than +7.3 kcal
(d) Zero
Can you answer this question?
 
 

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Answer: Less than +7.3 kcal
$\Delta H = \Delta E + \Delta n_g RT$
or $\Delta E = \Delta H - \Delta n_g RT$
$\therefore \Delta E = +7.3 - \frac{1}{2} \times 2 \times 298$ = (7300 - 298) cal = 7002 cal= 7.00 kcal
answered Feb 24, 2014 by mosymeow_1
 

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