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?

1 Answer
Truong-Son N.
Jul 10, 2016

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)#

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