CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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Chapter 5, Problem 5.148P

(a)

Interpretation Introduction

Interpretation:

The volumes of NH3 needed per liter of flue gas at 1 atm are to be calculated

Concept introduction:

The relationship between pressure and volume can be expressed as follows,

PV=constant

Here, P is the pressure and V is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

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.

(a)

Expert Solution
Check Mark

Answer to Problem 5.148P

The volumes of NH3 are needed per liter of flue gas at 1 atm are 4.5×105 L.

Explanation of Solution

The equation for the reaction of NH3, NO with O2 is as follows:

4NH3(g) + 4NO(g)+O2(g)N2(g)+6H2O(g) (1)

The formula to convert °C to Kelvin is:

TK=T°C+273.15 (2)

Substitute 365 °C for T°C in equation (2).

TK=365+273.15=638.15 K

The expression to calculate the moles of the NO is as follows,

PV=nRT (3)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the NO and R is the gas constant.

Rearrange the equation (3) to calculate n as follows,

n=PVRT                                                                                                                       (4)

Substitute the value 4.5×105 atm for P, 638.15 K for T, 1.0 L for V and 0.0821 LatmmolK for R in the equation (4).

n=(1.0 L)(4.5×105 atm)(0.0821 LatmmolK)(638.15 K)=8.589×107 mol

From the equation (1), four moles of NH3 reacts with four moles of NO so, the moles of NH3 is calculated from NO is as follows:

Moles of NH3=(Moles of NO)(2 mol of NH32 mol of NO) (5)

Substitute the value 8.589×107 mol for NO in the equation (5)

Moles of NH3=(8.589×107 mol)(2 mol of NH32 mol of NO)=8.589×107 mol

The expression to calculate the volume is as follows,

PV=nRT (3)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the NO and R is the gas constant.

Rearrange the equation (3) to calculate V as follows,

V=nRTP                                                                                                                     (6)

Substitute the value 1 atm for P, 638.15 K for T, 8.5891×107 mol for n and 0.0821 LatmmolK for R in the equation (6).

V=(8.5891×107 atm)(0.0821 LatmmolK)(638.15 K)(1.0 atm)=4.50001×105 L4.5×105 L

Conclusion

The volumes of NH3 are needed per liter of flue gas at 1 atm are 4.5×105 L.

(b)

Interpretation Introduction

Interpretation:

The mass of NH3 are needed per kiloliter at 365 °C and 1 atm is to be calculated.

Concept introduction:

The relationship between pressure and volume can be expressed as follows,

PV=constant

Here, P is the pressure and V is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

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.

(b)

Expert Solution
Check Mark

Answer to Problem 5.148P

The mass of NH3 are needed per kiloliter at 365 °C and 1 atm is 0.015 g.

Explanation of Solution

Substitute the value 4.5×105 atm for P, 638.15 K for T, 1.0 kL for V and 0.0821 LatmmolK for R in the equation (4).

n=(1.0 kL(1000 L1 kL))(4.5×105 atm)(0.0821 LatmmolK)(638.15 K)=8.589×104 mol

From the equation (1), four moles of NH3 reacts with four moles of NO so, the moles of NH3 is calculated from NO is as follows:

Moles of NH3=(Moles of NO)(2 mol of NH32 mol of NO) (7)

Substitute the value 8.589×104 mol for NO in the equation (7)

Moles of NH3=(8.589×104 mol)(2 mol of NH32 mol of NO)=8.589×104 mol

The expression to calculate the mass of NH3 is as follows:

Moles of NH3=Mass of NH3Molar mass of NH3 (8)

Rearrange the equation (8) to calculate the mass of NH3 as follows,

Mass of NH3=(Moles of NH3)(Molar mass of NH3)                                              (9)

Substitute the value 8.589×104 mol for moles of NH3 and 17.03 g/mol for the molar mass of NH3 in the equation (9).

Mass of NH3=(8.589×104 mol)(17.03 g/mol)=0.014627 g=0.015 g

Conclusion

The mass of NH3 are needed per kiloliter at 365 °C and 1 atm is 0.015 g.

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Chapter 5 Solutions

CHEMISTRY >CUSTOM<

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