Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.156P
Interpretation Introduction

Interpretation:

The partial pressure of I2 gas is to be calculated.

Concept introduction:

Ideal gas law relates pressure, volume, temperature, and number of moles of gas.

The ideal gas equation can be expressed as follows,

  PV=nRT        (1)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

The expression to calculate the partial pressure of the gas is as follows:

  Pgas=XgasPtotal

Here, Pgas is the partial pressure of the gas, Xgas mole fraction of the gas and Ptotal is the total pressure.

The expression to calculate the mole fraction of the gas is as follows:

  Xgas=Moles of gasTotal moles

Here, Xgas is the mole fraction of the gas.

Expert Solution & Answer
Check Mark

Answer to Problem 5.156P

The partial pressure of I2 gas is 25.2 torr.

Explanation of Solution

The balanced chemical equation for the reaction of I2 with F2 is as follows:

  7F2(g) + I2(s)2IF7(g)        (2)

Convert pressure from torr to atm is as follows:

  P=350 torr(1 atm760 torr)=0.460526315 atm

The expression to calculate the moles of the F2 is as follows:

  PV=nRT        (2)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of F2 and R is the gas constant.

Rearrange equation (1) for n.

  n=PVRT        (3)

Substitute 0.460526315 atm for P, 250K for T, 2.50 L for V and 0.0821 LatmmolK for R in equation (3) to calculate moles of F2.

  n=(0.460526315 atm)(2.50 L)(0.0821 LatmmolK)(250 K)=0.056093339 mol

The expression to calculate the moles of I2 is as follows:

  Moles of I2=Mass of I2Molar mass of I2        (4)

Substitute 2.50 g for the mass of I2 and 253.8g/mol for molar mass of I2 in the equation (4).

  Moles of I2=2.50 g253.8g/mol=0.009850 mol

According to the balanced chemical equation (2), seven moles of F2 yield two moles of IF7 hence, moles of IF7 produced from F2 is calculated as follows:

  Moles of IF7=[(Moles of F2)(2 mol I27 mol F2)]        (5)

Substitute 0.056093339 mol for moles of F2 in equation (5).

  Moles of IF7=[(0.056093339 mol)(2 mol IF77 mol F2)]=0.016026668mol

From the equation (2), one mole of I2 yields two moles of IF7 hence, moles of IF7 produced from I2 is calculated as follows:

  Moles of IF7=[(Moles of I2)(2 mol IF71 mol I2)]        (6)

Substitute the value 0.009850 mol for moles of I2 in the equation (6).

  Moles of IF7=[(0.009850 mol)(2 mol IF71 mol I2)]=0.019700551 mol

F2 is the limiting reagent because it produces less number of moles of IF7.

The stoichiometric ratio between I2 and F2 is 1:7. I2 is excess reagent and thus the amount of I2 reacted is calculated as,

  moles of I2 react with F2=(moles of F2)(1 mol I27 mol F2)        (7)

Substitute 0.056093339 mol for moles of F2 in equation (7).

  moles of I2 react with F2=(0.056093339 mol)(1 mol I27 mol F2)=0.00801333414 mol I2

The expression to calculate the mole I2 left after the reaction is as follows:

  Moles of I2=[(initial moles of I2)(moles of I2 react with F2)]        (8)

Substitute the value 0.009850 mol for initial moles of I2 and 0.0080133414mol for moles of I2 reacted with F2 in the equation (8).

  Moles of I2=[(0.009850 mol)(0.0080133414mol)]=1.83694×103 mol

The total moles of the gas after the reaction is completed is equal to the sum of moles of I2, F2 and IF7.

  Total moles=(moles of F2)+(moles of I2)+(moles of IF7)=(0 mol)+(1.83694×103 mol)+(0.019700551 mol)=0.0178636mol

Rearrange the equation (1) for P as follows:

  V=nRTP        (9)

Substitute 0.0178636mol for n, 550 K for T, 2.50 mL for V and 0.0821 LatmmolK for R in equation (9) to calculate the total pressure.

  P=(0.0178636mol)(0.0821 LatmmolK)(550 K)(2.50 L)=0.322652 atm

Convert pressure from atm to torr as follows:

  P=0.322652 atm(760 torr1 atm)=245.21552 torr245 torr

The expression to calculate the mole fraction of I2 is as follows:

  Xiodine=Moles of iodinetotal moles        (10)

Substitute 1.83694×103 for moles of iodine and 0.0178636mol for total moles in the equation (10).

  Xiodine=1.83694×103 mol0.0178636mol=0.102831

The expression to calculate the partial pressure of I2 is as follows:

  Piodine=(Xiodine)(Ptotal)        (11)

Substitute 0.102831 for Xiodine and 245.21552torr for Ptotal in the equation(11).

  Piodine=(0.102831)(245.21552torr)=25.215869torr25.2 torr.

Conclusion

The partial pressure of I2 gas is 25.2 torr.

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Chapter 5 Solutions

Chemistry: The Molecular Nature of Matter and Change

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