When 1 mol CS2 (l) forms from its elements at 1 atm and 25°C, 89.7 kJ of heat is absorbed, and it takes 27.7 kJ to vaporize 1 mol of the liquid. How much heat is absorbed when 1 mol CS2 (g) forms from its elements at these conditions

There is 117.4 kJ of heat absorbed

Explanation:

Step 1: Data given

Number of moles CS2 = 1 mol

Temperature = 25° = 273 + 25 = 298 Kelvin

Heat absorbed = 89.7 kJ

It takes 27.7 kJ to vaporize 1 mol of the liquid

Step 2: Calculate the heat that is absorbed

C (s) + 2S (s) → CS2 (l) ΔH = 89.7 kJ (positive since heat is absorbed)

CS2 (l) → CS2 (g) ΔH = 27.7 kJ (positive since heat is absorbed)

We should balance the equations, before summing, but since they are already balanced, we don't have to change anything.

C (s) + 2S (s) - --> CS2 (g)

ΔH = 89.7 + 27.7 = 117.4 kJ

There is 117.4 kJ of heat absorbed