Question

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?

Answer 1

It's important to realize that the `"5.66 kJ"` of heat released is actually `q`, the heat flow, not the enthalpy 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)`