   # Use Hess’s law to calculate the enthalpy change for the formation of CS 2 ( ℓ ) from C(s) and S(s) [C(s) + 2 S(s) → CS 2 ( ℓ )] from the following enthalpy values. C(s) + O 2 (g) → CO 2 (g) Δ r H 1 ∘ = − 393.5 kJ/mol-rxn S(s) + O 2 (g) → SO 2 (g) Δ r H 2 ∘ = − 296.8 kJ/mol-rxn CS 2 ( l ) + 3 O 2 (g) → CO 2 (g) + 2 SO 2 (g) Δ r H 3 ∘ = − 1103.9 kJ/mol-rxn ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 5.7, Problem 1CYU
Textbook Problem
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## Use Hess’s law to calculate the enthalpy change for the formation of CS2(ℓ) from C(s) and S(s) [C(s) + 2 S(s) → CS2(ℓ)] from the following enthalpy values. C(s) + O 2 (g) → CO 2 (g)  Δ r H 1 ∘ = − 393.5  kJ/mol-rxn S(s) + O 2 (g) → SO 2 (g)  Δ r H 2 ∘ = − 296.8  kJ/mol-rxn CS 2 ( l ) + 3 O 2 (g) → CO 2 (g) + 2 SO 2 (g)  Δ r H 3 ∘ = − 1103.9  kJ/mol-rxn

Interpretation Introduction

Interpretation:

The enthalpy change for the formation of CS2 has to be calculated.

Concept Introduction:

Hess’s law:

If a reaction proceeds in two or more reactions,ΔrH0 for the overall process is the sum of ΔrH0 values of all those reactions

### Explanation of Solution

The required equation for the formation of CS2 is the reverse of 3rd

Equation i.e,

CS2+3O2CO2+2SO2

ΔrH30=ΔrH30 =+1103.9kJ/mol

2nd equation is written for 2mol of SO2, i.e,

2S2+2O22SO2

ΔrH20=rH20=-296

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