Question

Consider the following process
$\Delta H\left(kJ/mol\right)$
$\frac{1}{2}A\to B+50$
$3B\to 3C+D-125$
$E+A\to 2D+350$
For $B+D\to E+2C,\Delta H$will be:

• A. 325 kJ/mol
• B. 525 kJ/mol
• C. -375 kJ/mol
• D. -325 kJ/mol

Answer 1

The correct option is C -375 kJ/mol

Solution:-

$\frac{1}{2}A⟶B+50$

$2\times [\frac{1}{2}A⟶B+50]$

$A⟶2B+100.....\left(1\right)$

$3B⟶2C+D-125.....\left(2\right)$

$E+A⟶2D+350$

$2D⟶E+A-350.....\left(3\right)$

Adding equation $\left(1\right),\left(2\right)\&\left(3\right)$, we have

$A+3B+2D⟶2B+100+2C+D-125+E+A-350$

$B+D⟶E+2C-375$

Hence the value of $\Delta H$ will be $-375KJ/mol$.