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Lady bird [3.3K]
3 years ago
14

Given the following equations, determine the standard enthalpy of formation (ΔH∘f) for one mole of ICl3(g).I2(g)+3Cl2(g)⟶2ICl3(g

)ΔH∘298=−214 kJI2(s)⟶I2(g)ΔH∘298=38 kJNote that iodine (I2) is a solid at standard conditions.Your answer should have two significant figures.
Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
6 0

The standard enthalpy of formation (ΔH°f) for one mole of ICl₃(g) is -88 kJ, as determined by Hess' law.

We want to determine the standard enthalpy of formation (ΔH°f) for one mole of ICl₃(g). The equation for which we are looking the enthalpy of reaction is:

0.5 I₂(s) + 1.5 Cl₂(g) ⟶ ICl₃(g)

We will use Hess' law, which states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken. Let's consider the following thermochemical equations.

I₂(g) + 3 Cl₂(g) ⟶ 2 ICl₃(g)   ΔH°298 = −214 kJ

I₂(s) ⟶ I₂(g)                           ΔH°298 = 38 kJ

We will add the reactions and their enthalpies. The resulting reaction is:

I₂(s) + 3 Cl₂(g) ⟶ 2 ICl₃(g)   ΔH°298 = −-176 kJ

Finally, since we want to calculate the standard enthalpy per mole of ICl₃, we will divide the previous equation by 2.

0.5 I₂(s) + 1.5 Cl₂(g) ⟶ ICl₃(g)   ΔH°298 = −88 kJ

The standard enthalpy of formation (ΔH°f) for one mole of ICl₃(g) is -88 kJ, as determined by Hess' law.

Learn  more: brainly.com/question/22953190

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