Question

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

Answer 1

There is 117.4 kJ of heat absorbed

Step-by-step 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