Question

A chemist mixes $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{g}}\mathbf{}{\mathbf{\text{CuCl}}}_{\mathbf{2}}$ $\mathbf{1}$with an excess of${\left({\text{NH}}_4\right)}_{\mathbf{2}}{\mathbf{\text{HPO}}}_{\mathbf{4}}$ in dilute aqueous solution. He measures the evolution of $\mathbf{670}\mathbf{}\mathbf{\text{J}}$heat as the two substances react to give ${\mathbf{\text{Cu}}}_{\mathbf{3}}{\left({\text{PO}}_4\right)}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{.}$Compute$\mathit{\Delta }\mathit{H}$ that would result from the reaction of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{\text{molCuCl}}}_{\mathbf{2}}$with an excess of${\left({\text{NH}}_4\right)}_{\mathbf{2}}{\mathbf{\text{HPO}}}_{\mathbf{4}}\mathbf{.}$

Short answer

The enthalpy change in reaction, $(∆H)=-90.1\text{k}J$.

Step-by-step solution

Given data

The enthalpy change$∆H=-670\text{J}/\text{g.}$

The number of moles in $\text{CuCl}2$is $1.00 \text{mol}$.

Concept of the enthalpy change

The heat change that occurs when a chemical reaction occurs at constant volume or pressure is known as enthalpy change. The quantity of heat absorbed or released during the reaction is determined by the enthalpy change. The letter $\mathit{\Delta }\mathit{H}$ stands for it.

Calculation of heat change  (ΔH)

The calculation of the molar mass of copper chloride is shown below.

$Mr\left(CuC{l}_2\right)=134.5\frac{\text{g}}{\text{mol}}$

Now, calculate the enthalpy change.

$$\begin{gathered} \Delta H=1.00{\text{mol CuCl}}_2\times \frac{134.5\text{g}\text{CuCl}}{1\text{mol}{\text{CuCl}}_2}\times \frac{-670\text{J}}{1.00\text{g}{\text{CuCl}}_2} \\ =-90.1\times {10}^3\text{J} \\ =-90.1\text{kJ} \end{gathered}$$

Therefore, the enthalpy changes of the reaction $\left(\Delta H\right)=-90.1\text{k}J$.