Question #05692

1 Answer
Ernest Z.
Jan 24, 2016

You are almost correct. #ΔH_"con" = "-23.3 kJ/mol"#

Explanation:

#"NH"_3("l") → "NH"_3("g"); ΔH "= 68.5 kJ"#

#"Moles of NH"_3 = 50.0 color(red)(cancel(color(black)("g"))) × "1 mol"/(17.0 color(red)(cancel(color(black)("g")))) = "2.941 mol"#

#ΔH_"vap" = "68.5 kJ"/"2.941 mol" = "23.3 kJ/mol"#

#ΔH_"con" = -ΔH_"vap" = "-23.3 kJ/mol"#

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