Use the appropriate table to calculate \(\Delta H^{\circ}\) for (a) the reaction between copper(II) oxide and carbon monoxide to give copper metal and carbon dioxide. (b) the decomposition of one mole of methyl alcohol (CH \(_{3} \mathrm{OH}\) ) to methane and oxygen gases.

Short Answer

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Question: Calculate the standard enthalpy change (ΔH°) for the following reactions: (a) CuO (s) + CO (g) -> Cu (s) + CO2 (g) (b) 2 CH3OH (l) -> CH4 (g) + O2 (g) + H2O (l) Answer: (a) ΔH° = -125.7 kJ/mol (b) ΔH° = 116.8 kJ/mol

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for this reaction is: CuO (s) + CO (g) -> Cu (s) + CO2 (g)
02

Find the standard enthalpies of formation

According to a table of standard enthalpies of formation, we have: ΔHf°(CuO) = -157.3 kJ/mol ΔHf°(CO) = -110.5 kJ/mol ΔHf°(Cu) = 0 kJ/mol (Standard enthalpy of formation for elements in their standard states is always zero) ΔHf°(CO2) = -393.5 kJ/mol
03

Calculate ΔH° for the reaction

Use the equation mentioned in the analysis: ΔH° = Σ ΔHf°(products) - Σ ΔHf°(reactants) ΔH° = [ΔHf°(Cu) + ΔHf°(CO2)] - [ΔHf°(CuO) + ΔHf°(CO)] ΔH° = [(0) + (-393.5)] - [(-157.3) + (-110.5)] ΔH° = -393.5 - (-267.8) ΔH° = -125.7 kJ/mol So, the standard enthalpy change for this reaction is ΔH° = -125.7 kJ/mol. (b) Decomposition of one mole of methyl alcohol to methane and oxygen gases:
04

Write the balanced chemical equation

The balanced chemical equation for this reaction is: 2 CH3OH (l) -> CH4 (g) + O2 (g) + H2O (l)
05

Find the standard enthalpies of formation

According to a table of standard enthalpies of formation, we have: ΔHf°(CH3OH) = -238.7 kJ/mol ΔHf°(CH4) = -74.8 kJ/mol ΔHf°(O2) = 0 kJ/mol (Standard enthalpy of formation for elements in their standard states is always zero) ΔHf°(H2O) = -285.8 kJ/mol
06

Calculate ΔH° for the reaction

Use the equation mentioned in the analysis: ΔH° = Σ ΔHf°(products) - Σ ΔHf°(reactants) ΔH° = [ΔHf°(CH4) + ΔHf°(O2) + ΔHf°(H2O)] - [2 × ΔHf°(CH3OH)] ΔH° = [(-74.8) + (0) + (-285.8)] - [2 × (-238.7)] ΔH° = -360.6 - (-477.4) ΔH° = 116.8 kJ/mol So, the standard enthalpy change for this reaction is ΔH° = 116.8 kJ/mol.

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Most popular questions from this chapter

In the late eighteenth century Priestley prepared ammonia by reacting \(\mathrm{HNO}_{3}(g)\) with hydrogen gas. The thermodynamic equation for the reaction is $$ \mathrm{HNO}_{3}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-637 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) when one mole of hydrogen gas reacts. (b) What is \(\Delta H\) when \(10.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g)\) is made to react with an excess of steam to form \(\mathrm{HNO}_{3}\) and \(\mathrm{H}_{2}\) gases?

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