   Chapter 13, Problem 99CP

Chapter
Section
Textbook Problem

# A 1.604-g sample of methane (CH4) gas and 6.400 g oxygen gas are scaled into a 2.50-L vessel at 411°C and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium. the pressure of oxygen is 0.326 atm, and the pressure of water vapor is 4.45 atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

Interpretation Introduction

Interpretation: A given mass of a sample of methane gas and gas oxygen is given to be present sealed in a 2.50L vessel at 411°C . These are allowed to reach the equilibrium state. The equilibrium pressure of oxygen and water vapor is given, the equilibrium pressures of carbon monoxide and carbon dioxide are to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction.

To determine: The equilibrium pressures of carbon monoxide and carbon dioxide.

The number of moles of oxygen gas present initially is 0.2mol_ .

Explanation

Explanation

Given

The two reactions are,

CH4(g)+O2(g)CO2(g)+2H2O(g)

CH4(g)+32O2(g)CO(g)+2H2O(g)

The initial mass of CH4 gas is 1.604g .

The initial mass of O2 gas is 6.400g .

The volume of the vessel is 2.50L .

The equilibrium pressure of oxygen is 0.326atm .

The equilibrium pressure of water vapor is 4.45atm .

The molar mass of O2 =2O=2(16)g/mol=32g/mol

The number of moles of a substance is calculated by the formula,

Numberofmoles=GivenmassMolarmass

Substitute the value of the given mass and the molar mass of oxygen in the above expression.

Numberofmoles=6.400g32g/mol=0.2mol_

The change in the partial pressure of oxygen gas is 4.16atm_ .

Given

The temperature is 411°C(684K) .

The number of moles of oxygen gas present initially is 0.2mol .

The initial partial pressure of the oxygen gas is calculated by the ideal gas equation,

POV=nOxygenRTPO=nOxygenRTV

Where,

• PO is the initial partial pressure of oxygen.
• V is the volume o the vessel.
• nOxygen is the initial number of moles of oxygen.
• R is the universal gas constant (0.0821Latmmol1K1) .
• T is the temperature.

Substitute the value of V , nOxygen , R and T in the above expression.

PO=nOxygenRTV=(0.2mol)(0.0821Latmmol1K1)(684K)(2.50L)=4.49atm

The change in the partial pressure of oxygen is calculated by the formula,

Change=(InitialpartialpressureofO2EqilibriumpartialpressureofO2)

Substitute the value of the initial and the equilibrium pressure of O2 in the above expression.

Change=(4.490.3126)atm=4

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