Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Chapter 3, Problem 3C.9E

(a)

Interpretation Introduction

Interpretation:

Volume of hydrogen needed to produce 1.0 t NH3 has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  PV=nRT

Here,

P is pressure of the gas.

V is volume of gas.

n denotes moles of a gas.

R is gas constant.

T is temperature of gas.

(a)

Expert Solution
Check Mark

Answer to Problem 3C.9E

Volume of hydrogen needed to produce 1.0 t NH3 is 3.0×105 L.

Explanation of Solution

Balanced reaction of Haber process is as follows:

  N2+3H22NH3

Quantity of moles of 1.0 t NH3 can be calculated as follows:

  Mole(NH3)=Given massMolar mass=(1.0 t17.03 g/mol)(103 kg1 t)(103 g1 kg)=5.9×104 mol

According to reaction, 3 moles of H2 gas produces 2 moles of NH3 thus amount of H2 required to produce 5.9×104 mol of NH3 can be calculated as follows:

  Mass(H2)=Mole(NH3)(3 mol H22 mol NH3)=(5.9×104 mol NH3)(3 mol H22 mol NH3)=8.8×104 mol H2

The expression to calculate volume of gas is as follows:

  V=nRTP        (1)

Conversion factor to convert 350°C to Kelvin is as follows:

  T(K)=temperatureincelsius+273.15=350°C+273.15=623.15K

Substitute 15.00 atm for P, 8.8×104 mol for n, 8.20574×102 LatmK1mol1 for R and 623.15K for T in equation (1).

  V=(8.8×104 mol)(8.20574×102 LatmK1mol1)(623.15K)(15.00 atm)=3.0×105 L

Hence, volume of hydrogen needed to produce 1.0 t NH3 is 3.0×105 L.

(b)

Interpretation Introduction

Interpretation:

Volume of hydrogen needed to produce 1.0 t NH3 at 376 atm and 250 °C has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  PV=nRT

Here,

P is pressure of the gas.

V is volume of gas.

n denotes moles of a gas.

R is gas constant.

T is temperature of gas.

(b)

Expert Solution
Check Mark

Answer to Problem 3C.9E

Volume of hydrogen needed to produce 1.0 t NH3 at 376 atm and 250 °C is 1.0×104 L.

Explanation of Solution

The expression to calculate volume of gas is as follows:

  V=nRTP        (1)

Conversion factor to convert 250°C to Kelvin is as follows:

  T(K)=temperatureincelsius+273.15=250°C+273.15=523.15K

Substitute 376 atm for P, 8.8×104 mol for n, 8.20574×102 LatmK1mol1 for R and 523.15K for T in equation (1).

  V=(8.8×104 mol)(8.20574×102 LatmK1mol1)(523.15K)(376 atm)=1.0×104 L

Hence volume of hydrogen needed to produce 1.0 t NH3 at 376 atm and 250 °C is 1.0×104 L.

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Chapter 3 Solutions

Chemical Principles: The Quest for Insight

Ch. 3 - Prob. 3A.5ECh. 3 - Prob. 3A.6ECh. 3 - Prob. 3A.7ECh. 3 - Prob. 3A.8ECh. 3 - Prob. 3A.9ECh. 3 - Prob. 3A.10ECh. 3 - Prob. 3B.1ASTCh. 3 - Prob. 3B.1BSTCh. 3 - Prob. 3B.2ASTCh. 3 - Prob. 3B.2BSTCh. 3 - Prob. 3B.3ASTCh. 3 - Prob. 3B.3BSTCh. 3 - Prob. 3B.4ASTCh. 3 - Prob. 3B.4BSTCh. 3 - Prob. 3B.5ASTCh. 3 - Prob. 3B.5BSTCh. 3 - Prob. 3B.6ASTCh. 3 - Prob. 3B.6BSTCh. 3 - Prob. 3B.7ASTCh. 3 - Prob. 3B.7BSTCh. 3 - Prob. 3B.8ASTCh. 3 - Prob. 3B.8BSTCh. 3 - Prob. 3B.1ECh. 3 - Prob. 3B.2ECh. 3 - Prob. 3B.5ECh. 3 - Prob. 3B.6ECh. 3 - Prob. 3B.9ECh. 3 - Prob. 3B.10ECh. 3 - Prob. 3B.11ECh. 3 - Prob. 3B.12ECh. 3 - Prob. 3B.13ECh. 3 - Prob. 3B.14ECh. 3 - Prob. 3B.15ECh. 3 - Prob. 3B.16ECh. 3 - Prob. 3B.17ECh. 3 - Prob. 3B.18ECh. 3 - Prob. 3B.19ECh. 3 - Prob. 3B.20ECh. 3 - Prob. 3B.21ECh. 3 - Prob. 3B.22ECh. 3 - Prob. 3B.23ECh. 3 - Prob. 3B.24ECh. 3 - Prob. 3B.25ECh. 3 - Prob. 3B.26ECh. 3 - Prob. 3B.27ECh. 3 - Prob. 3B.28ECh. 3 - Prob. 3B.29ECh. 3 - Prob. 3B.30ECh. 3 - Prob. 3B.31ECh. 3 - Prob. 3B.32ECh. 3 - 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Prob. 3G.7ECh. 3 - Prob. 3G.8ECh. 3 - Prob. 3G.9ECh. 3 - Prob. 3G.10ECh. 3 - Prob. 3G.11ECh. 3 - Prob. 3G.12ECh. 3 - Prob. 3G.13ECh. 3 - Prob. 3G.14ECh. 3 - Prob. 3G.15ECh. 3 - Prob. 3G.16ECh. 3 - Prob. 3G.17ECh. 3 - Prob. 3G.18ECh. 3 - Prob. 3H.1ASTCh. 3 - Prob. 3H.1BSTCh. 3 - Prob. 3H.2ASTCh. 3 - Prob. 3H.2BSTCh. 3 - Prob. 3H.3ASTCh. 3 - Prob. 3H.3BSTCh. 3 - Prob. 3H.4ASTCh. 3 - Prob. 3H.4BSTCh. 3 - Prob. 3H.5ASTCh. 3 - Prob. 3H.5BSTCh. 3 - Prob. 3H.1ECh. 3 - Prob. 3H.2ECh. 3 - Prob. 3H.3ECh. 3 - Prob. 3H.4ECh. 3 - Prob. 3H.5ECh. 3 - Prob. 3H.6ECh. 3 - Prob. 3H.7ECh. 3 - Prob. 3H.8ECh. 3 - Prob. 3H.9ECh. 3 - Prob. 3H.10ECh. 3 - Prob. 3H.11ECh. 3 - Prob. 3H.12ECh. 3 - Prob. 3H.13ECh. 3 - Prob. 3H.14ECh. 3 - Prob. 3H.15ECh. 3 - Prob. 3H.16ECh. 3 - Prob. 3H.17ECh. 3 - Prob. 3H.19ECh. 3 - Prob. 3H.20ECh. 3 - Prob. 3H.23ECh. 3 - Prob. 3H.24ECh. 3 - Prob. 3H.25ECh. 3 - Prob. 3H.26ECh. 3 - Prob. 3H.27ECh. 3 - Prob. 3H.28ECh. 3 - Prob. 3H.29ECh. 3 - Prob. 3H.30ECh. 3 - Prob. 3H.31ECh. 3 - Prob. 3H.32ECh. 3 - 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Prob. 3.4ECh. 3 - Prob. 3.5ECh. 3 - Prob. 3.6ECh. 3 - Prob. 3.7ECh. 3 - Prob. 3.8ECh. 3 - Prob. 3.9ECh. 3 - Prob. 3.10ECh. 3 - Prob. 3.11ECh. 3 - Prob. 3.12ECh. 3 - Prob. 3.13ECh. 3 - Prob. 3.15ECh. 3 - Prob. 3.18ECh. 3 - Prob. 3.19ECh. 3 - Prob. 3.23ECh. 3 - Prob. 3.24ECh. 3 - Prob. 3.25ECh. 3 - Prob. 3.26ECh. 3 - Prob. 3.27ECh. 3 - Prob. 3.29ECh. 3 - Prob. 3.31ECh. 3 - Prob. 3.32ECh. 3 - Prob. 3.35ECh. 3 - Prob. 3.36ECh. 3 - Prob. 3.37ECh. 3 - Prob. 3.38ECh. 3 - Prob. 3.40ECh. 3 - Prob. 3.41ECh. 3 - Prob. 3.42ECh. 3 - Prob. 3.45ECh. 3 - Prob. 3.47ECh. 3 - Prob. 3.49ECh. 3 - Prob. 3.50ECh. 3 - Prob. 3.51ECh. 3 - Prob. 3.53ECh. 3 - Prob. 3.54ECh. 3 - Prob. 3.55ECh. 3 - Prob. 3.56ECh. 3 - Prob. 3.57ECh. 3 - Prob. 3.58ECh. 3 - Prob. 3.59ECh. 3 - Prob. 3.60ECh. 3 - Prob. 3.61ECh. 3 - Prob. 3.62ECh. 3 - Prob. 3.63ECh. 3 - Prob. 3.64ECh. 3 - Prob. 3.65ECh. 3 - Prob. 3.66ECh. 3 - Prob. 3.67ECh. 3 - Prob. 3.68E
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