Question

When 1.00 g potassium chlorate $\left({\mathrm{KClO}}_3\right)$ is dissolved in 50.0 g water in a Styrofoam calorimeter of negligible heat capacity, the temperature decreases from 25.00°C to 23.36°C. Calculate qfor the water and${\mathbf{\Delta H}}^{\mathbf{o}}$ for the process.

data-custom-editor="chemistry" ${\mathbf{KClO}}_{\mathbf{3}}\left(\mathbf{s}\right)\to {\mathbf{K}}^{\mathbf{+}}\left(\mathbf{aq}\right){\mathbf{+}\mathbf{ClO}}_{\mathbf{3}}^{\mathbf{-}}\left(\mathbf{aq}\right)$

The specific heat of water is $\mathbf{4}\mathbf{.}\mathbf{184}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$.

Short answer

The value of${\mathrm{q}}_{{\mathrm{H}}_2\mathrm{O}}$and${\mathrm{\Delta H}}^{\mathrm{o}}$ is 343.1 J.

Step-by-step solution

The given Information.

The mass of potassium chlorate is 1.00 g.

The mass of water is 50.0 g.

The initial temperature is $23.36°\mathrm{C}$.

The final temperature is $25.00°\mathrm{C}$.

The specific heat of water is $4.184{\mathrm{JK}}^{-1}{\mathrm{g}}^{-1}$.

Heat and enthalpy change.

The heat taken up or given out by a particular reaction at constant pressure is equivalent to enthalpy change. It is designated by the symbol $\mathrm{\Delta H}$.

Calculating the q of water and the ΔHo of the process.

The value of q can be found as:

$\mathrm{q}=\mathrm{mC\Delta T}$

Here, m is the mass of the substance; C is the specific heat capacity and ΔT is the change in temperature.

Substituting the values in the equation,

$\begin{array}{c}\mathrm{q}=\mathrm{mC\Delta T}\\ =50\mathrm{g}\times \frac{18\mathrm{J}}{{}^{\circ }\mathrm{C}}\times \left(25-23.6\right){}^{\circ }\mathrm{C}\\ =343.088\mathrm{J}\end{array}$

A Styrofoam calorimeter is a constant-pressure calorimeter.

The q is equal to the enthalpy change at constant pressure.

Therefore,

$\mathrm{\Delta H}=343.088\mathrm{J}$

Hence, the value of${\mathrm{q}}_{{\mathrm{H}}_2\mathrm{O}}$ is 343.1 J and the value${\mathrm{\Delta H}}^{\mathrm{o}}$ of is 343.1 J.