Question

Consider the following reaction,
$2A+B\to C+2D$, $\Delta H_1=10$
$A+2C\to 2D+B$, $\Delta H_2=-5$ What is $\Delta H$ of reaction $A+2B\to 3C$?

• A. $-5$
• B. $+5$
• C. $+10$
• D. $+15$

Answer 1

Answer: (D)

$+15$

Solution:- (D) $+15$

$A+2B⟶3C\Delta H=?$

Given:-

$2A+B⟶C+2D\Delta H_1=10.....\left(1\right)$

$A+2C⟶2D+B\Delta H_2=-5$

$B+2D⟶A+2C\Delta H_3=-\Delta H_2=5.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

$2A+B+B+2D⟶C+2D+A+2C\Delta H=10+5$

$A+2B⟶3C\Delta H=15$

Hence the $\Delta H$ for the given reaction is $+15$.