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Wittaler [7]
3 years ago
6

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

st of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)
Chemistry
1 answer:
Anastasy [175]3 years ago
7 0

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

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

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

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−153.1 J / (K mol)

Explanation:

Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)

Hg(liq) + Cl2(g) → HgCl2(s)

Given that;

The standard molar entropies of the species at that temperature are:

Sºm (Hg,liq) = 76.02 J / (K mol) ;

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Sºm (HgCl2,s) = 146.0 J / (K mol)

The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]

= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]

= -153.09 J / (K mol)

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