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Ket [755]
3 years ago
5

Using standard heats of formation, calculate the standard enthalpy change for the following reaction.

Chemistry
1 answer:
In-s [12.5K]3 years ago
6 0

Answer:

ΔH°r = -114.40 kJ

Explanation:

Let's consider the following balanced equation.

4 HCl(g) + O₂(g) ⇒ 2 H₂O(g) + 2 Cl₂(g)

We can calculate the standard enthalpy change of the reaction (ΔH°r) using the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where

  • ΔH°f: standard heat of formation
  • n: moles
  • p: products
  • r: reactants

ΔH°r = 2 mol × ΔH°f(H₂O(g)) + 2 mol × ΔH°f(Cl₂(g)) - 4 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(O₂(g))

ΔH°r = 2 mol × (-241.82 kJ/mol) + 2 mol × 0 kJ/mol - 4 mol × (-92.31 kJ/mol) - 1 mol × 0 kJ/mol

ΔH°r = -114.40 kJ

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Answer:

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You don't give "the data below", so I will use my best estimates from the NIST Chemistry WebBook. You can later substitute your own values.

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C₂ = specific heat capacity of liquid   = 115 J·K⁻¹mol⁻¹

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ΔT₁ = 78.4 - 300 = -221.6 K

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q = -15.60 - 53.4 - 4.802 = -50.6 kJ

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