# A sample of ammonia liberates 5.66 kJ of heat as it solidifies at its melting point. What is the mass of the sample? DeltaH_(solid)  = -5.66 kJ/mol?

Jul 10, 2016

It's important to realize that the $\text{5.66 kJ}$ of heat released is actually $q$, the heat flow, not the enthalpy of freezing, $\Delta {H}_{\text{frz}}$.

"Solidification at the melting point" is otherwise known as freezing at the freezing point, so this process occurs at ${T}_{f} = - {77.73}^{\circ} \text{C}$ for ammonia.

The equation that relates enthalpy to heat flow at constant pressure is:

$\setminus m a t h b f \left(\Delta {\overline{H}}_{\text{frz" = q_p/n_"compound}}\right)$

where:

• $\Delta {\overline{H}}_{\text{frz}}$ is the molar enthalpy of freezing, in $\text{kJ/mol}$.
• ${q}_{p}$ is the heat flow $q$ at a constant pressure.
• ${n}_{\text{compound}}$ is the $\text{mol}$s of the compound that is freezing.

So, all you need to realize is that you know ${q}_{p}$ and $\Delta {H}_{\text{frz}}$. Therefore:

n_"compound" = q_p/(DeltaH_"frz")

$= \left(- \text{5.66 kJ")/(-"5.66 kJ/mol}\right)$

$=$ $\textcolor{b l u e}{\text{1 mol}}$

Thus, the mass of ammonia is:

= "1 mol" xx (14.007 + 3xx1.007"9 g NH"_3)/("1 mol NH"_3)

$= \textcolor{b l u e}{{\text{17.0307 g NH}}_{3}}$