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Chapter 6: Problem 69

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 \(-283.3 \mathrm{~kJ} /\) 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 \( 2 C(s) + O_{2}(g) \rightarrow 2 CO(g) \) is \( +172.9 \: kJ/mol \).

Step by step solution

01

Write down the given reactions and their enthalpy changes

The given reactions are: 1. \( C(s) + O_{2}(g) \rightarrow CO_{2}(g) \) , \(\Delta H_{1} = -393.7 \: kJ/mol \) 2. \( CO(g) + \frac{1}{2} O_{2}(g) \rightarrow CO_{2}(g) \), \(\Delta H_{2} = -283.3 \: kJ/mol \)
02

Manipulate the given reactions to match the target reaction

We need to rearrange and combine these reactions to match the given reaction: \( 2 C(s) + O_{2}(g) \rightarrow 2 CO(g) \). First, we will reverse the second reaction: \( CO_{2}(g) \rightarrow CO(g) + \frac{1}{2} O_{2}(g) \), \(\Delta H'_{2} = +283.3 \: kJ/mol \) (enthalpy change is negated when reversing a reaction) Now, we need to multiply this reversed reaction by 2 to match the coefficients of the target reaction: \( 2 CO_{2}(g) \rightarrow 2 CO(g) + O_{2}(g) \), \(\Delta H''_{2} = 2 \times (+283.3 \: kJ/mol) = +566.6 \: kJ/mol \)
03

Combine the manipulated reactions

Now, we can combine the given reactions and their modified enthalpy changes to obtain the target reaction: First Reaction: \( C(s) + O_{2}(g) \rightarrow CO_{2}(g) \) , \(\Delta H_{1} = -393.7 \: kJ/mol \) Modified Second Reaction (multiplied by 2 and reversed): \( 2 CO_{2}(g) \rightarrow 2 CO(g) + O_{2}(g) \), \(\Delta H''_{2} = +566.6 \: kJ/mol \) Adding these two reactions, we get: \( 2 C(s) + O_{2}(g) \rightarrow 2 CO(g) \)
04

Calculate ΔH for the target reaction

We can now calculate the total enthalpy change for the given reaction by adding the enthalpy changes of the manipulated reactions: \(\Delta H = \Delta H_{1} + \Delta H''_{2} = (-393.7 \: kJ/mol) + (+566.6 \: kJ/mol) = +172.9 \: kJ/mol \) Thus, the enthalpy change for the reaction \( 2 C(s) + O_{2}(g) \rightarrow 2 CO(g) \) is \( +172.9 \: kJ/mol \).

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Most popular questions from this chapter

Explain the advantages and disadvantages of hydrogen as an alternative fuel.

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

The overall reaction in a commercial heat pack can be represented as $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad \Delta H=-1652 \mathrm{~kJ} $$ a. How much heat is released when \(4.00\) moles of iron are reacted with excess \(\mathrm{O}_{2}\) ? b. How much heat is released when \(1.00\) mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is produced? c. How much heat is released when \(1.00 \mathrm{~g}\) iron is reacted with excess \(\mathrm{O}_{2}\) ? d. How much heat is released when \(10.0 \mathrm{~g} \mathrm{Fe}\) and \(2.00 \mathrm{~g} \mathrm{O}_{2}\) are reacted?

The enthalpy change for a reaction is a state function and it is an extensive property. Explain.

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5 \mathrm{~L}\). Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.
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