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Chapter 7: Problem 67

The enthalpy of combustion of solid carbon to form carbon dioxide is \(-393.7 \mathrm{kJ} / \mathrm{mol}\) carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is \(-283.3 \mathrm{kJ} / \mathrm{mol}\) CO. Use these data to calculate \(\Delta H\) for the reaction $$2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g)$$

Short Answer

Expert verified
The enthalpy change for the reaction \(2C(s) + O_{2}(g) \rightarrow 2CO(g)\) is \(\Delta H = 220.8 \mathrm{kJ/mol}\).

Step by step solution

01

Write the given reactions and enthalpy changes

We have the following reactions and the corresponding enthalpy changes: 1. C(s) + O2(g) → CO2(g) ΔH1 = -393.7 kJ/mol C 2. CO(g) + 1/2 O2(g) → CO2(g) ΔH2 = -283.3 kJ/mol CO Our goal is to find the enthalpy change (ΔH) for the reaction: 2C(s) + O2(g) → 2CO(g).
02

Manipulate the given reactions to align with the target reaction

First, we need to determine how the given reactions can be combined or manipulated to match the target reaction. To get 2CO(g) as a product, we should modify the second reaction: Given Reaction: CO(g) + 1/2 O2(g) → CO2(g) ΔH2 = -283.3 kJ/mol CO Multiply the reaction by 2: New Reaction: 2CO(g) + O2(g) → 2CO2(g) ΔH2' = 2(-283.3 kJ/mol CO) = -566.6 kJ Now, we should manipulate the first reaction to have CO2(g) as a reactant: Given Reaction: C(s) + O2(g) → CO2(g) ΔH1 = -393.7 kJ/mol C Reverse the reaction: New Reaction: CO2(g) → C(s) + O2(g) ΔH1' = -(-393.7 kJ/mol C) = 393.7 kJ/mol C
03

Use Hess's Law to calculate the enthalpy change of the target reaction

Now we can use Hess's Law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of the individual reactions. Add the modified reactions from Step 2 to get the target reaction: 2CO(g) + O2(g) → 2CO2(g) ΔH2' = -566.6 kJ 2CO2(g) → 2C(s) + 2O2(g) ΔH1' = 2(393.7 kJ/mol C) = 787.4 kJ/mol C Sum: 2C(s) + O2(g) → 2CO(g) ΔH = ΔH1' + ΔH2'
04

Calculate the enthalpy change for the target reaction

Now we will add the enthalpy changes of the modified reactions: ΔH = ΔH1' + ΔH2' ΔH = 787.4 kJ + (-566.6 kJ) ΔH = 220.8 kJ/mol The enthalpy change for the reaction 2C(s) + O2(g) → 2CO(g) is ΔH = 220.8 kJ/mol.

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Most popular questions from this chapter

In a coffee-cup calorimeter, \(150.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl is added to \(50.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) to make \(200.0 \mathrm{g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\). If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{kJ} / \mathrm{mol},\) calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is \(4.184 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) and assume no heat loss to the surroundings.

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