Chapter 10, Problem 10.121QP

Interpretation Introduction

Interpretation: The change in enthalpy on molar basis has to be calculated.

Concept Introduction:

Specific heat can be defined as quantity of heat required to raise the temperature of $\text{1}\text{g}$ substance by $\text{1°C}$. The relationship between heat and change in temperature can be expressed by the equation given below.

$\text{q=cmΔT}$

Where $\text{q=}$ Heat added

c= Specific heat

$\text{m=}$ Mass

$\text{ΔT=}$ Change in temperature.

The unit of specific heat is $\text{J}{\text{g}}^{\text{-1}}\text{.°C}$ .

Answer to Problem 10.121QP

The change in standard enthalpy is

                              $\begin{array}{l}\text{-3}{\text{.60×10}}^{\text{2}}\text{kJ}{\text{mol}}^{\text{-1}}\text{Zn}\\ \text{(or)}\text{-3}\text{.60-3}{\text{.60×10}}^{\text{2}}\text{kJ}{\text{2mol}}^{\text{-1}}{\text{Ag}}^{\text{+}}\end{array}$

Explanation of Solution

Record the given data

Volume and Molarity of Silver Nitrate = $\text{50}\text{.0}\text{mL}\text{\&}\text{0}\text{.100}\text{M}$

Initial temperature = $\text{19}\text{.25°C}$

Final temperature = $\text{22}\text{.17°C}$

Heat capacity of calorimeter = $\text{98}\text{.6}\text{J}\text{per}\text{°C}$

To calculate the heat produced

Heat produced by the reaction = ${\text{q}}_{\text{reaction}}{\text{=-(q}}_{\text{solution}}{\text{+q}}_{\text{cal}}\text{)}$

${\text{q}}_{\text{solution}}\text{=msΔT}$ = $\text{(50}\text{.0}\text{g)(4}\text{.184}\text{J}{\text{g}}^{\text{-1}}\text{°C)(22}\text{.17-19}\text{.25)°C}$

= $\text{611}\text{J}$

${\text{q}}_{\text{cal}}\text{=CΔT=(98}\text{.6}\text{J}\text{per°C)(22}\text{.17-19}\text{.25)°C=288}\text{J}$

$\begin{array}{l}{\text{-q}}_{\text{reaction}}{\text{=(q}}_{\text{solution}}{\text{+q}}_{\text{cal}}\text{)}\\ \\ \text{=(611+288)J}\\ \\ \text{=899}\text{J}\end{array}$

Heat produced by the reaction = $\text{899}\text{J}$

To calculate the heat produced on molar basis

$\text{899}\text{J}$ was produced from $\text{50}\text{.0}\text{mL}\text{of}\text{0}\text{.100}\text{M}$ Silver nitrate solution

Moles of Silver = $\text{50}\text{.0}\text{mL×}\frac{\text{0}\text{.100}\text{mol}{\text{Ag}}^{\text{+}}}{\text{1000}\text{mL}\text{soln}}\text{ = 5}{\text{.00×10}}^{\text{-3}}\text{mol}$

On molar basis, heat produced

= $\frac{\text{899}\text{J}}{\text{5}{\text{.00×10}}^{\text{-3}}\text{mol}{\text{Ag}}^{\text{+}}}\text{=1}{\text{.80×10}}^{\text{5}}\text{J}{\text{mol}}^{\text{-1}}{\text{Ag}}^{\text{+}}\text{=180}\text{kJ}{\text{mol}}^{\text{-1}}{\text{Ag}}^{\text{+}}$

Heat produced = $\text{180}\text{kJ}{\text{mol}}^{\text{-1}}$

To calculate the enthalpy of the reaction

The equation shows 2 moles of Silver, therefore the heat produced =

$\text{2}\text{mol×180}\text{kJ}{\text{mol}}^{\text{-1}}\text{=360}\text{kJ}$

Heat is produced by the reaction, then,

$\begin{array}{l}{\text{ΔH}}_{\text{reaction}}{\text{=q}}_{\text{reaction}}\text{=-3}{\text{.60×10}}^{\text{2}}\text{kJ}{\text{mol}}^{\text{-1}}\text{Zn}\\ \text{(or)}\\ {\text{ΔH}}_{\text{reaction}}{\text{=q}}_{\text{reaction}}\text{=-3}{\text{.60×10}}^{\text{2}}\text{kJ}{\text{2mol}}^{\text{-1}}{\text{Ag}}^{\text{+}}\end{array}$

Conclusion

The enthalpy change of the reaction was calculated using the values of heat produced by moles of Silver.  The heat produced by the reaction was found to be $\text{-3}{\text{.60×10}}^{\text{2}}\text{kJ}{\text{mol}}^{\text{-1}}\text{Zn}\text{(or)}\text{-3}{\text{.60×10}}^{\text{2}}\text{kJ}{\text{2mol}}^{\text{-1}}{\text{Ag}}^{\text{+}}$