# 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?

It’s important to realize that the ##”5.66 kJ”## of heat released is actually ##q##, the heat flow, not the of freezing, ##DeltaH_”frz”##.

“Solidification at the melting point” is otherwise known as freezing at the freezing point, so this process occurs at ##T_f = -77.73^@ “C”## for ammonia.

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

##mathbf(DeltabarH_”frz” = q_p/n_”compound”)##

where:

• ##DeltabarH_”frz”## is the molar enthalpy of freezing, in ##”kJ/mol”##.
• ##q_p## is the heat flow ##q## at a constant pressure.
• ##n_”compound”## is the ##”mol”##s of the compound that is freezing.

So, all you need to realize is that you know ##q_p## and ##DeltaH_”frz”##. Therefore:

##n_”compound” = q_p/(DeltaH_”frz”)##

##= (-“5.66 kJ”)/(-“5.66 kJ/mol”)##

##=## ##color(blue)(“1 mol”)##

Thus, the mass of ammonia is:

##= “1 mol” xx (14.007 + 3xx1.007″9 g NH”_3)/(“1 mol NH”_3)##

##= color(blue)(“17.0307 g NH”_3)##

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