Chapter EN, Problem 17PQ

Interpretation Introduction

Interpretation:

$\Delta H^{\text{o}}$ of following reaction has to be determined.

  ${\text{2CrO}}_{{}_4}^{2-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

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 17PQ

Option (C) is the correct one.

Explanation of Solution

Reason for correct option:

The given reaction is as follows:

  ${\text{2CrO}}_{{}_4}^{2-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

$\text{2}\text{mol}$ of ${\text{CrO}}_{{}_4}^{2-}$ reacts with $2\text{mol}$ of ${\text{H}}^{+}$ to form $1\text{mol}$ of ${\text{Cr}}_2{\text{O}}_7^{2-}$ and $1\text{mol}$ of ${\text{H}}_2\text{O}$.

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

  $\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\Delta H_{\text{f}}^{°}\left[{\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)\right]+\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(l\right)\right]\right\}-\\ \left\{2\Delta H_{\text{f}}^{°}\left[{\text{CrO}}_{{}_4}^{2-}\left(aq\right)\right]+2\Delta H_{\text{f}}^{°}\left[{\text{H}}^{+}\left(aq\right)\right]\right\}\end{array}\right]$        (1)

Substitute $-\text{1490}\text{.3 kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)\right]$, $-285.8\text{ kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}_2\text{O}\left(l\right)\right]$, $-881.2\text{ kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{CrO}}_{{}_4}^{2-}\left(aq\right)\right]$ and $0\text{ kJ/mol}$ for $\Delta H_{\text{f}}^{°}\left[{\text{H}}^{+}\left(aq\right)\right]$ in equation (1).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left\{\left(-\text{1490}\text{.3 kJ/mol}\right)+\left(-285.8\text{ kJ/mol}\right)\right\}-\\ \left\{2\left(-881.2\text{ kJ/mol}\right)+2\left(0\text{ kJ/mol}\right)\right\}\end{array}\right]\\ =-13.7\text{kJ/mol}\end{array}$

$\Delta H^{\text{o}}$ of following reaction is $-13.7\text{kJ/mol}$, hence correct option is (C).

  ${\text{2CrO}}_{{}_4}^{2-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

Reason for incorrect option:

Since $\Delta H^{\text{o}}$ of following reaction is $-13.7\text{kJ/mol}$,

  ${\text{2CrO}}_{{}_4}^{2-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

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

Conclusion

$\Delta H^{\text{o}}$ of following reaction is $-13.7\text{kJ/mol}$, hence correct option is (C).

  ${\text{2CrO}}_{{}_4}^{2-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Cr}}_2{\text{O}}_7^{2-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$