Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 5, Problem 135AP

Acetylene   ( C 2 H 2 ) can be made by combining calcium carbide ( CaC 2 ) with water. (a) Write an equation for the reaction. (b) What is the maximum amount of beat (in Joules) that can be obtained from the combustion of acetylene, starting with 74 .6g of CaC 2 ?

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Interpretation Introduction

Interpretation:

The balanced equation for the given reaction is to be written and the maximum amount of heat that can be obtained from the combustion of acetylene is to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy change which occurs at standard conditions.

The standard enthalpy of a reaction is to be determined by using the equation as given below:

ΔH°rxn= nΔHf°(products) mΔHf°(reactants)

Here, the stoichiometric coefficients are represented by m for the reactants and n for the products. The enthalpy of formation at standard conditions is represented by ΔHf°.

The value of enthalpy of formation of an element is zero at its most stable state.

In one joule, there are 0.001 kJ.

Conversion factor for J to kJ is 0.001 kJ1 J.

Answer to Problem 135AP

Solution:

(a) CaC2(s)+2 H2O(l)Ca(OH)2(s)+C2H2(g)

(b) 1.51×106 J

Explanation of Solution

a) An equation for reaction.

The reaction of calcium carbide with water in order to form acetylene is as follows:

CaC2(s)+H2O(l)Ca(OH)2(s)+C2H2(g)

Now, balance this equation to obtain the balanced chemical equation as shown below:

CaC2(s)+2 H2O(l)Ca(OH)2(s)+C2H2(g)

Hence, the balanced chemical equation is:

CaC2(s)+2 H2O(l)Ca(OH)2(s)+C2H2(g).

Given information: Mass, mCaC2=74.6 g.

b) The maximum amount of heat that can be obtained from the combination of acetylene, starting with 74.6 g of CaC2.

The combustion of acetylene takes place as follows:

2 C2H2(g)+5 O2(g)4 CO2(g)+2 H2O(l)

The enthalpy of formation of oxygen gas is zero because it is in the most stable form.

From appendix 2,

ΔH°f(C2H2)=226.6 kJ/molΔH°f(H2O)=285.8 kJ/molΔH°f(CO2)=393.5 kJ/mol

Calculate the standard enthalpy of reaction as follows:

ΔH°rxn=[ 4ΔH°f(CO2)+2 ΔH°f(H2O) ][ 2ΔH°f(C2H2)+5 ΔH°f(O2) ]

Substitute 226.6 kJ/mol for ΔH°f(C2H2), 285.8 kJ/mol for ΔH°f(H2O), 393.5 kJ/mol for ΔH°f(CO2), and 0 for ΔH°f(O2) in the above equation and solve:

ΔH°rxn=[ 4(393.5 kJ/mol)+2(285.8 kJ/mol) ][ 2(226.6 kJ/mol)+5(0) ]=2599 kJ/mol

This enthalpy is required for the formation of 2 mol

of C2H2.

The molar mass of CaC2

is 64 g/mol

Th mass of CaC2 is 74.6 g

Calculate the number of moles as follows:

nCaC2=mCaC2MCaC2

Substitute 74.6 g for mCaC2 and 64 g/mol for MCaC2 in the above equation.

nCaC2=74.6 g64 g/mol=1.16 mol

One mol of C2H2 is produced from one mol of CaC2, and therefore the number of moles of C2H2

is equal to number of moles of CaC2.

nC2H2=1.16 mol

Calculate the amount of heat required when 1.16 mol of acetylene is burned as follows:

(1.16 mol C2H2)(2599 kJ2 mol C2H2 )=1.51×103 kJ

In one kilojoule, there are 1000 J.

Convert kilojoule to joule as follows:

1.51×103 kJ=(1.51×103 kJ)(1000 J1 kJ)=1.51×106 J

Hence, the heat produced from combustion of acetylene is 1.51×106 J

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Chapter 5 Solutions

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