Chemistry - Thermodynamics - II Question with Solution | TestHub

To calculate $\Delta H_{400\mathrm{K}}^{\circ }$ for the reaction $2{\mathrm{A}}_{(\mathrm{g})}+3{\mathrm{B}}_{(\mathrm{g})}\to 2{\mathrm{C}}_{(\mathrm{g})}+{\mathrm{D}}_{(\mathrm{g})}$, we use the Gibbs-Helmholtz equation.

First, calculate $\Delta G_T^{\circ }$ for the reaction at temperature $T$:

$$\Delta G_T^{\circ }=\sum (\nu \Delta G_f^{\circ }{)}_{\mathrm{products}}-\sum (\nu \Delta G_f^{\circ }{)}_{\mathrm{reactants}}$$

Where $\nu$ represents the stoichiometric coefficients.

Given:

$\Delta G_f^{\circ }\mathrm{A}=(200+10^{-2}{T}^2)\mathrm{J}$

$\Delta G_f^{\circ }\mathrm{B}=(100+2\times 10^{-3}{T}^2)\mathrm{J}$

$\Delta G_f^{\circ }\mathrm{C}=(200+2\times 10^{-2}{T}^2)\mathrm{J}$

$\Delta G_f^{\circ }\mathrm{D}=(100+10^{-3}{T}^2)\mathrm{J}$

Substitute these into the equation for $\Delta G_T^{\circ }$:

$$\Delta G_T^{\circ }=[2(200+2\times 10^{-2}{T}^2)+1(100+10^{-3}{T}^2)]-[2(200+10^{-2}{T}^2)+3(100+2\times 10^{-3}{T}^2)]$$

$$\Delta G_T^{\circ }=[400+4\times 10^{-2}{T}^2+100+10^{-3}{T}^2]-[400+2\times 10^{-2}{T}^2+300+6\times 10^{-3}{T}^2]$$

$$\Delta G_T^{\circ }=[500+(0.04+0.001)T^2]-[700+(0.02+0.006)T^2]$$

$$\Delta G_T^{\circ }=[500+0.041T^2]-[700+0.026T^2]$$

$$\Delta G_T^{\circ }=-200+0.015T^2\mathrm{J}$$

Now, use the Gibbs-Helmholtz equation, which states that at constant pressure:

$${\left[\frac{\partial (\Delta G/T)}{\partial T}\right]}_P=-\frac{\Delta H}{T^2}$$

Rearranging for $\Delta H$:

$$\Delta H=-T^2{\left[\frac{\partial (\Delta G/T)}{\partial T}\right]}_P$$

First, calculate $\Delta G/T$:

$$\frac{\Delta G_T^{\circ }}{T}=\frac{-200+0.015T^2}{T}=-\frac{200}{T}+0.015T$$

Next, differentiate $\frac{\Delta G_T^{\circ }}{T}$ with respect to $T$:

$$\frac{\partial (\Delta G_T^{\circ }/T)}{\partial T}=\frac{\partial }{\partial T}(-200T^{-1}+0.015T)$$

$$=-200(-1)T^{-2}+0.015$$

$$=\frac{200}{T^2}+0.015$$

Now, substitute this back into the $\Delta H$ equation:

$$\Delta H_T^{\circ }=-T^2\left(\frac{200}{T^2}+0.015\right)$$

$$\Delta H_T^{\circ }=-200-0.015T^2\mathrm{J}$$

Finally, calculate $\Delta H^{\circ }$ at $T=400\mathrm{K}$:

$$\Delta H_{400\mathrm{K}}^{\circ }=-200-0.015(400{)}^2$$

$$\Delta H_{400\mathrm{K}}^{\circ }=-200-0.015(160000)$$

$$\Delta H_{400\mathrm{K}}^{\circ }=-200-2400$$

$$\Delta H_{400\mathrm{K}}^{\circ }=-2600\mathrm{J}$$

The question asks for the magnitude in kJ, rounded to the nearest integer.

Magnitude $=|-2600\mathrm{J}|=2600\mathrm{J}$

Convert to kJ: $2600\mathrm{J}=2.6\mathrm{kJ}$

Rounding to the nearest integer, the magnitude is $3\mathrm{kJ}$.