Textbook Question
Consider the following equilibrium, for which at 𝐾𝑝 = 0.0752 at 480°C: 2 Cl2(𝑔) + 2 H2O(𝑔) ⇌ 4 HCl(𝑔) + O2(𝑔) (a) What is the value of 𝐾𝑝 for the reaction 4 HCl(𝑔) + O2(𝑔) ⇌ 2 Cl2(𝑔) + 2 H2O(𝑔)?
8. Thermochemistry
Hess's Law
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Calculate ΔHrxn for the following reaction using Hess's Law: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g). Given the following reactions and their ΔH values: 1) 2Fe(s) + 3/2O2(g) → Fe2O3(s), ΔH = -824.2 kJ; 2) CO(g) + 1/2O2(g) → CO2(g), ΔH = -282.7 kJ.
A
-197.9 kJ
B
-564.6 kJ
C
-1124.2 kJ
D
-1406.9 kJ
Verified step by step guidance1
Understand Hess's Law: It states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in. This means you can add up the enthalpy changes of individual steps to find the overall enthalpy change.
Identify the target reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g). We need to manipulate the given reactions to match this target reaction.
Reverse the first given reaction: 2Fe(s) + 3/2O2(g) → Fe2O3(s), ΔH = -824.2 kJ. Reversing it gives Fe2O3(s) → 2Fe(s) + 3/2O2(g), and the ΔH becomes +824.2 kJ.
Multiply the second given reaction by 3: CO(g) + 1/2O2(g) → CO2(g), ΔH = -282.7 kJ. Multiplying by 3 gives 3CO(g) + 3/2O2(g) → 3CO2(g), and the ΔH becomes 3 × (-282.7 kJ) = -848.1 kJ.
Add the modified reactions: Combine the reversed first reaction and the multiplied second reaction. The O2 terms will cancel out, leaving the target reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g). Add the ΔH values: +824.2 kJ + (-848.1 kJ) to find ΔHrxn.
