Question

A $\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{-}\mathbf{g}$ chunk of sodium metal is cautiously dropped into a mixture of $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{g}$of water and $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{g}$ of ice, both at $\mathbf{0}\mathbf{°}\mathbf{C}$. The reaction is

$\mathbf{2}\mathbf{Na}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{NaOH}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_2\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\Delta H}\mathbf{=}\mathbf{-}\mathbf{368}\mathbf{kJ}$

Will the ice melt? Assuming the final mixture has a specific heat capacity of $\mathbf{4}\mathbf{.}\mathbf{18}{\mathbf{Jg}}^{-1}{}^o\mathbf{C}^{-1}\mathbf{,}$calculate the final temperature.

Short answer

No, the ice will not melt as the energy removed by sodium hydroxide is not enough. The final temperature of the final mixture is $0°\mathrm{C}$.

Step-by-step solution

Determine the energy removed by sodium

Energy removed:

$\begin{array}{c}\mathrm{Q}=250\mathrm{g}\times \left(\frac{1\mathrm{mol}}{23\mathrm{g}}\right)\times \left(\frac{368\mathrm{kJ}}{2\mathrm{mol}}\right)\\ =2.00\mathrm{kJ}\end{array}$

Determine the energy acquired for melting ice

The energy requiredto melt the ice of mass $50\mathrm{g}$ is

$\begin{array}{c}{\mathrm{Q}}_1=50\mathrm{g}\times \frac{1\mathrm{mol}}{18.02}\times \frac{6.02\mathrm{kJ}}{1\mathrm{mol}}\\ =16.7\mathrm{kJ}\end{array}$

Since the energy removed is not present in enough amount to melt the ice, the final temperature of the mixture will remain at $0^{\circ }\mathrm{C}$.