Question

14.4 g sample of liquid ammonia is heated at a constant pressure of 1 atm from a temperature of 239.8 K to a temperature of 324.9 K. How much energy in kJ is required?

Answer 1

You must supply 10.59 kJ.

Explanation:

Here is a schematic heating curve for ammonia.

The boiling point of ammonia is 239.8 K (-33.4 °C).

Thus, you are starting with liquid ammonia at Point B on the graph, heating the liquid at its boiling point until it has wholly vaporized at Point C, and then heating the vapour until the temperature reaches 324.9 K (51.8 °C) at Point D.

There are two separate heat inputs involved in this problem:

• `q_1` = heat added to vaporize the ammonia at 239.8 K
• `q_2` = energy added to heat the vapour from 239.8 K to 324.9 K

`q = q_1 + q_2 = nΔ_text(vap)H + mC_text(s)ΔT`

where

`"ncolor(white)(mmm) =` the moles of ammonia

`Δ_text(vap)Hcolor(white)(l) =` the molar enthalpy of vaporization of ammonia (+5.653 kJ/mol)

`mcolor(white)(mmll) =` the mass of the ammonia

`C_text(s)color(white)(mml) =`the specific heat capacity of gaseous ammonia (`"4.744 J·K"^"-1""g"^"-1"`)

`ΔT color(white)(mm)= T_"f" -T_"i"`

`bbq_1`

`n = 14.4 color(red)(cancel(color(black)("g"))) × "1 mol"/(17.03 color(red)(cancel(color(black)("g")))) = "0.8456 mol"`

`q_1 = nΔ_text(vap)H = 0.8456 color(red)(cancel(color(black)("mol"))) × ("+5.653 kJ"·color(red)(cancel(color(black)("mol"^"-1")))) = "+4.780 kJ"`

`bbq_2`

`ΔT = "324.9 K - 239.8 K = 85.1 K"`

`q_2 = mC_text(s)ΔT = 14.4 color(red)(cancel(color(black)("g"))) × 4.744 color(white)(l)"J"·color(red)(cancel(color(black)( "K"^"-1""g"^"-1"))) × 85.1 color(red)(cancel(color(black)("K"))) = "5813 J = 5.813 kJ"`

`q = q_1 + q_2 = "(4.780 + 5.813) kJ = 10.59 kJ"`

You must supply 10.59 kJ.