Chapter EN, Problem 14PQ

Interpretation Introduction

Interpretation:

Whether standard enthalpy of formation of $\text{AgCl}$ is more exothermic or endothermic form $\text{AgBr}$ has to be determined.

Concept Introduction:

Standard enthalpy of reaction is calculated by summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactants at the standard conditions. The expression to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum n\Delta H_{\text{f (reactants)}}^{°}$

Here,

  m is the stoichiometric coefficient of the product.

  n is the stoichiometric coefficient of reactant.

  $\Delta H_{\text{f}\left(\text{reactants}\right)}^{°}$ is the standard enthalpy of reactant formation.

  $\Delta H_{\text{f}\left(\text{products}\right)}^{°}$ is the standard enthalpy of product formation.

Answer to Problem 14PQ

Option (A) is the correct one.

Explanation of Solution

Reason for correct option:

The formation of $\text{AgCl}$ from its elements is as follows:

  $\text{Ag}\left(s\right)+\frac{1}{2}{\text{Cl}}_2\left(g\right)\to \text{AgCl}\left(s\right)$

One mole of $\text{Ag}$ reacts with half moles of ${\text{Cl}}_2$ to form one mole of $\text{AgCl}$.

The formula to calculate standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\Delta H_{\text{f}}^{°}\left[\text{AgCl}\left(s\right)\right]\right\}-\\ \left\{\Delta H_{\text{f}}^{°}\left[\text{Ag}\left(s\right)\right]+\frac{1}{2}\Delta H_{\text{f}}^{°}\left[{\text{Cl}}_2\left(g\right)\right]\right\}\end{array}\right]$        (1)

$\Delta H_{\text{f}}^{°}\left[{\text{Cl}}_2\left(g\right)\right]$ has 0 value as ${\text{Cl}}_2$ exist as diatomic gas in its most stable form and $\Delta H_{\text{f}}^{°}\left[\text{Ag}\left(s\right)\right]$ has 0 value as $\text{Ag}$ exist as solid in its stable form.

Substitute $\text{0 kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{Cl}}_2\left(g\right)\right]$, $0\text{ kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[\text{Ag}\left(s\right)\right]$, $-\text{127 kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[\text{AgCl}\left(s\right)\right]$ in equation (1).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\left\{-\text{127 kJ/mol}\right\}-\left\{\left(0\text{kJ/mol}\right)+\frac{1}{2}\left(0\text{kJ/mol}\right)\right\}\right]\\ =-\text{127 kJ/mol}\end{array}$

The formation of $\text{AgBr}$ from its elements is as follows:

  $\text{Ag}\left(s\right)+\frac{1}{2}{\text{Br}}_2\left(g\right)\to \text{AgBr}\left(s\right)$

One mole of $\text{Ag}$ reacts with half mole of ${\text{Br}}_2$ to form one mole of $\text{AgBr}$.

The formula to calculate standard enthalpy of a given reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

  $\Delta H_{\text{rxn}}^{°}=\left[\left\{\Delta H_{\text{f}}^{°}\left[\text{AgBr}\left(s\right)\right]\right\}-\left\{\Delta H_{\text{f}}^{°}\left[\text{Ag}\left(s\right)\right]+\frac{1}{2}\Delta H_{\text{f}}^{°}\left[{\text{Br}}_2\left(g\right)\right]\right\}\right]$        (2)

$\Delta H_{\text{f}}^{°}\left[{\text{Br}}_2\left(g\right)\right]$ has 0 value as ${\text{Br}}_2$ exist as diatomic gas in its most stable form and $\Delta H_{\text{f}}^{°}\left[\text{Ag}\left(s\right)\right]$ has 0 value as $\text{Ag}$ exist as solid in its stable form.

Substitute $\text{0 kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{Br}}_2\left(g\right)\right]$, $0\text{ kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[\text{Ag}\left(s\right)\right]$, $-\text{100}\text{.4 kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[\text{AgBr}\left(s\right)\right]$ in equation (2).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\left\{-\text{100}\text{.4 kJ/mol}\right\}-\left\{\left(0\text{kJ/mol}\right)+\frac{1}{2}\left(0\text{kJ/mol}\right)\right\}\right]\\ =-\text{100}\text{.4 kJ/mol}\end{array}$

The standard enthalpy of formation of $\text{AgCl}$ is more exothermic than that for $\text{AgBr}$. Hence, correct option is (A).

Reason for incorrect option:

Since enthalpy of $\text{AgCl}$$\left(-\text{127 kJ/mol}\right)$ is less negative than for $\text{AgBr}$$\left(\text{100}\text{.4 kJ/mol}\right)$ so standard enthalpy of formation of $\text{AgCl}$ is more exothermic but not endothermic or less exothermic than that for $\text{AgBr}$.

Therefore, choice (C), (B) and (D) are incorrect.

Conclusion

The standard enthalpy of formation of $\text{AgCl}$ is less endothermic than that for $\text{AgBr}$. Hence, correct option is (A).