Question

For the hypothetical reaction, $A_2\left(g\right)+B_2\left(g\right)\rightleftharpoons 2AB\left(g\right)$ ${\Delta }_r{G}^{\circ }$ and ${\Delta }_r{S}^{\circ }$ are $20\text{kJ/mol}$ and $-20{\text{JK}}^{-1}{\text{mol}}^{-1}$ respectively at 200 K. If ${\Delta }_r{C}_p$ is $20{\text{JK}}^{-1}{\text{mol}}^{-1}$ then ${\Delta }_r{H}^{\circ }$ at 400K is:

• A. $20.0\text{kJ/mol}$
• B. $7.98\text{kJ/mol}$
• C. $28.0\text{kJ/mol}$
• D. $4.00\text{kJ/mol}$

Answer 1

The correct option is A $20.0\text{kJ/mol}$

Given data:

${\Delta }_r{C}_p^0=20J{K}^{-1}/mol$

$at=200K$

${\Delta }_r{G}^0=20kJ/mol$

${\Delta }_r{S}^0=-20J/mol$

$at=400$

For calculating change in enthalpy of a reaction we are applying Kirchoff`s equation:

$\Delta H_{T_2}^0=\Delta H_{T_1}^0+{\int }_{T_1}^{T_2}{C}_{p}dT$

First we calculate ${\Delta }_r^0{H}_{200K}$:

We know that

${\Delta }_r{G}^0={\Delta }_r^0{H}_{200K}-T{\Delta }_r{S}^0$
$20={\Delta }_r^0{H}_{200K}-\frac{200\times (-20)}{1000}$

${\Delta }_r{H}_{200K}^0=16kJ$

On substituting all values in kirchoff`s equation

$\Delta H_{T_{400}}^0=16+\frac{{\int }_{200}^{400}20}{1000}dT$

$\Delta H_{T_{400}}^0=16+\frac{20\times 200}{1000}=20.0kJ/mol$