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When 1 mol of 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 of CS₂(g) forms from its elements at these conditions?

Respuesta :

The heat absorbed when 1 mol of CS₂(g) forms from its elements at these conditions is 117.4 kJ.

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

Calculate the heat absorbed:

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

CS₂(l) →  CS₂g)           ΔH = 27.7 kJ  (positive since heat is absorbed)

Prior to summing, the equations should be balanced; however, because they are balanced already, there is nothing further we need to do.

C(s) + 2S(s)--->  CS₂ (g)

ΔH = 89.7 + 27.7 = 117.4 kJ

Hence, the heat absorbed is 117.4 kJ.

To know more about heat absorbed refer to: https://brainly.com/question/9588553

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