Chapter 22, Problem 80PQ

To determine

The heat absorbed or released in the reaction.

Answer to Problem 80PQ

The heat released in the reaction is $-3.1\times {10}^4\text{J}$.

Explanation of Solution

Write the expression to calculate the entropy of reactant A.

  $S_A={\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{c_{A}dT}{T}}$                                                                                                           (I)

Here, $S_A$ is the entropy of reactant A, $c_A$ is the molar specific heat capacity of A at constant pressure and T is the temperature.

Write the expression to calculate the entropy of reactant B.

  $S_B={\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{c_{B}dT}{T}}$                                                                                                          (II)

Here, $S_B$ is the entropy of reactant A and $c_B$ is the molar specific heat capacity of B at constant pressure.

Write the expression to calculate the entropy of reactant D.

  $S_D={\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{c_{D}dT}{T}}$                                                                                                        (III)

Here, $S_D$ is the entropy of reactant A, $c_D$ is the molar specific heat capacity of A at constant pressure.

Write the expression to calculate the entropy change for the reaction.

  $\Delta s=S_D-\left(S_A+S_B\right)$                                                                                               (IV)

Here, $\Delta s$ is the change in entropy of the reaction.

Write the expression to calculate the heat.

  $Q=T\Delta s$                                                                                                                (V)

Here, Q is the amount of heat.

Conclusion:

Substitute $2\sqrt{T}$ for $c_A$ in the above equation (I) to calculate $S_A$.

  $\begin{array}{c}{S}_A={\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{\left(2\sqrt{T}\right)dT}{T}}\\ =2{\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{dT}{\sqrt{T}}}\\ =2{\left[2\sqrt{T}\right]}_{0\text{K}}^{300\text{K}}\\ =69.3\text{J}/\text{K}\end{array}$

Substitute $4\sqrt{T}$ for $c_B$ in the above equation (II) to calculate $S_B$.

  $\begin{array}{c}{S}_A={\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{\left(4\sqrt{T}\right)dT}{T}}\\ =4{\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{dT}{\sqrt{T}}}\\ =4{\left[2\sqrt{T}\right]}_{0\text{K}}^{300\text{K}}\\ =138.6\text{J}/\text{K}\end{array}$

Substitute $3\sqrt{T}$ for $c_D$ in the above equation (III) to calculate $S_D$.

  $\begin{array}{c}{S}_A={\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{\left(3\sqrt{T}\right)dT}{T}}\\ =3{\displaystyle {\int }_{0\text{K}}^{300\text{K}}\frac{dT}{\sqrt{T}}}\\ =3{\left[2\sqrt{T}\right]}_{0\text{K}}^{300\text{K}}\\ =103.9\text{J}/\text{K}\end{array}$

Substitute $69.3\text{J}/\text{K}$ for $S_A$, $138.6\text{J}/\text{K}$ for $S_B$ and $103.9\text{J}/\text{K}$ for $S_D$ in the above equation (IV) to calculate $\Delta s$.

  $\begin{array}{c}\Delta s=103.9\text{J}/\text{K}-\left(69.3\text{J}/\text{K}+138.6\text{J}/\text{K}\right)\\ =-104\text{J}/\text{K}\end{array}$

Substitute $-104\text{J}/\text{K}$ for $\Delta s$ and $300\text{K}$ for T in the equation (V) to calculate Q.

  $\begin{array}{c}Q=300\text{K}\left(-104\text{J}/\text{K}\right)\\ =-3.1\times {10}^4\text{J}\end{array}$

The value of Q is the negative thus, the heat will release the system after the reaction.

Therefore, the heat released in the reaction is $-3.1\times {10}^4\text{J}$.