If #4.25 xx 10^3# #"g"# of gold was heated from #15.5^@ "C"# to #197.0^@ "C"#, how much heat was involved, and what is the molar enthalpy in #"kJ/mol"#? #C_P = "0.129 J/g"^@ "C"#?

Question

Natalie Anton · Accepted Answer

#"99.5 kJ"#, or #"4.61 kJ/mol"#, for what is apparently #"21.6 mols"# of gold... How rich is this person?

Well, we should look up the melting point of gold first. It's #1064^@ "C"#, so we can safely assume that only heating is involved (and not melting).

The heat flow equation is to be used at constant atmospheric pressure:

#q_P = mC_PDeltaT#,

where:

Thus, if we assume the specific heat capacity does not change in this temperature range, the heat absorbed is

#color(blue)(q_P) = (4.25 xx 10^3 "g Au")("0.129 J/g"^@ "C")(197.0 - 15.5^ @"C")#

#=# #9.95 xx 10^4# #"J"#

#=# #color(blue)("99.5 kJ")#

At constant pressure, the heat flow is also related to the molar enthalpy:

#DeltabarH = q_P/(n_"obj")#

where #n_"obj"# is just the mols of the object.

Therefore, the enthalpy for this heating process at constant pressure is:

#color(blue)(DeltabarH) = ("99.5 kJ")/(4.25 xx 10^3 cancel"g Au" xx "1 mol"/(196.96657 cancel"g Au"))#

#=# #color(blue)("4.61 kJ/mol")# to three sig figs.

Cameron Andrews · Answer

undefined The heat involved is 2.88 x 10^5 J, and the molar enthalpy is 64.6 kJ/mol.