Chapter 4, Problem 102QRT

Interpretation Introduction

Interpretation:

The ${\Delta }_r{H}^{\circ }$ of each step has to be calculated. The reaction enthalpy for overall process has to be identified and whether the given reaction is exothermic or not has to be indicated.

Concept Introduction:

Hess’s Law: The total enthalpy change for a chemical reaction is the summation of all changes in multiple stages of that chemical reaction.

Answer to Problem 102QRT

For Given Step 1, the value is $\text{-137}\text{.23 kJ}$.

For Given Step 2, the value is $\text{275}\text{.341 kJ}$.

For Given Step 3, the value is $\text{103}\text{.71 kJ}$.

The standard reaction enthalpy value for net equation is $241.82\text{ kJ}$.

The overall reaction is endothermic.

Explanation of Solution

Given Step 1:

  $S{O}_2\left(s\right)+2H_{2}O\left(g\right)+B{r}_2\left(g\right)\to H_{2}S{O}_4\left(l\right)+2HBr\left(g\right)$

Using the above given reaction enthalpy is calculated by subtracting the summation of formation of products from the summation of formation enthalpy of reactants.

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{ = }\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{l}\right){\text{+2Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{HBr}\left(\text{g}\right)\right]\text{-}\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{SO}}_{\text{2}}\left(\text{g}\right)\right]\text{-2}\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right]\text{-}\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{Br}}_{\text{2}}\left(\text{g}\right)\right]\\ \text{= }\left[\left(\text{-813}\text{.989kJ/mol}\right)\text{+2}\left(\text{-36}\text{.40 kJ/mol}\right)\right]\text{-}\left(\text{-296}\text{.830 kJ/mol}\right)\text{-2}\left(\text{-241}\text{.818 kJ/mol}\right)\\ \text{ -}\left(\text{-30}\text{.907 kJ/mol}\right)\\ =\text{ -137}\text{.23 kJ}\end{array}$

Given Step 2:

  $H_{2}S{O}_4\left(l\right)\to H_{2}O\left(g\right)+S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

Using the above given reaction enthalpy is calculated by subtracting the summation of formation of products from the summation of formation enthalpy of reactants.

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{ = }\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right){\text{+ Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{SO}}_{\text{2}}\left(\text{g}\right)\right]\text{-}\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{l}\right)\right]\\ \text{= }\left[\left(\text{-241}\text{.818kJ/mol}\right)\text{+}\left(\text{-296}\text{.830 kJ/mol}\right)\right]\text{-}\left(\text{- 813}\text{.989kJ/mol}\right)\\ =\text{ 275}\text{.341 kJ}\end{array}$

Given Step 3:

  $2HBr\left(g\right)\to H_2\left(g\right)+B{r}_2\left(g\right)$

Using the above given reaction enthalpy is calculated by subtracting the summation of formation of products from the summation of formation enthalpy of reactants.

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{ = }\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{\text{Br}}_{\text{2}}\left(\text{g}\right)\right]\text{-2}\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}\text{HBr}\left(\text{g}\right)\right]\\ \text{= }\left[\left(\text{30}\text{.907 kJ/mol}\right)\text{- 2}\left(\text{-36}\text{.40 kJ/mol}\right)\right]\\ =\text{ 103}\text{.71 kJ}\end{array}$

The above three expressions are added in order to obtain the net equation as follows,

Summation of the given equations and eliminating the reactants that show as products during summation are done. Then, the remaining reactant and products which form the net equation for the reaction and their individual reaction enthalpy changes values are added to get the net reaction enthalpy change value.

    $\begin{array}{l}S{O}_2\left(s\right)+2H_{2}O\left(g\right)+B{r}_2\left(g\right)\to H_{2}S{O}_4\left(l\right)+2HBr\left(g\right){\Delta }_r{H}^{\circ }=-137.23\text{ kJ}\\ H_{2}S{O}_4\left(l\right)\to H_{2}O\left(g\right)+S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right){\Delta }_r{H}^{\circ }=275.341\text{ kJ}\\ \underset{\_}{2HBr\left(g\right)\to H_2\left(g\right)+B{r}_2\left(g\right){\Delta }_r{H}^{\circ }=103.71kJ}\\ H_{2}O\left(g\right)\to H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right){\Delta }_r{H}^{\circ }=241.82\text{ kJ}\\ \end{array}$

The obtained net equation value is positive hence, the reaction is endothermic.