Question

Consider the following processes:
$\Delta H\left(kJ/mol\right)$
$\frac{1}{2}A\to B$ $+150$

$3B\to 2C+D$ $-125$

$E+A\to 2D$ $+350$

For $B+D\to E+2C,\Delta H$ will be:

• A. $525kJ/mol$
• B. $-175kJ/mol$
• C. $-325kJ/mol$
• D. $325kJ/mol$

Answer 1

The correct option is B $-175kJ/mol$

The processes are as shown below.
$\Delta H$
$\frac{1}{2}A⟶B$ $+150$ ............(1)
$3B⟶2C+D$ $-125$ ...............(2)
$E+A⟶2D$ $+350$ ..................(3)
________________________
$B+D⟶E+2C$ .........(4)

The reaction (1) is multiplied with 2 and added to the reaction (2). Then reaction (3) is subtracted to obtain reaction (4).

Hence, the expression for the enthalpy change for the reaction (4) is as given below.
$\Delta H_{\left(4\right)}=2\Delta H_{\left(1\right)}+\Delta H_{\left(2\right)}-\Delta H_{\left(3\right)}=300-125-350=-175$

Hence, for $B+D\to E+2C,\Delta H$ will be $-175kJ/mol$.