Question

The standard enthalpy change for the following reaction is -704 kJ at 298 K. Al(s) + 3/2 Cl2(g) — AICI3(8) AH° = -704 kJ WhatThe standard enthalpy change for the following reaction is 851 kJ at 298 K. 2 NaOH(s) —— 2 Na(s) + O2(g) + H2(g) AH° = 851 kJ

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Answer #1

1)

This reaction is -2*given reaction

So,

ΔHo rxn = -2*given ΔHo rxn

= -2*(-704 KJ)

= 1408 KJ

Answer: 1408 KJ

2)

This reaction is (-1/2)*given reaction

So,

ΔHo rxn = (-1/2)*given ΔHo rxn

= (-1/2)*(851 KJ)

= -425.5 KJ

Answer: -425.5 KJ

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