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Chapter 12: Problem 24

In the Haber process for the production of ammonia, $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

Short Answer

Expert verified
The relationship between the rate of production of ammonia (\(R_{NH_3}\)) and the rate of consumption of hydrogen (\(R_{H_2}\)) in the Haber process is given by: \( R_{NH_3}=\frac{2}{3} R_{H_2} \). This means that the rate of production of ammonia is 2/3 times the rate of consumption of hydrogen.

Step by step solution

01

Analyze the balanced chemical equation

For the Haber process, the balanced chemical equation is given by: \[ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \] The stoichiometry of the equation tells us that 1 mole of nitrogen (N₂) reacts with 3 moles of hydrogen (H₂) to produce 2 moles of ammonia (NH₃).
02

Express the rate of production of ammonia as a function of the rate of consumption of hydrogen

We can use the stoichiometry to write the relationship between the rate of production of ammonia (NH₃) and the rate of consumption of hydrogen (H₂). 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃, which means that the ratio of moles of NH₃ produced to moles of H₂ consumed is 2:3. Let's denote the rate of production of ammonia by \(R_{NH_3}\) and the rate of consumption of hydrogen by \(R_{H_2}\). Therefore, the relationship between the rate of production of ammonia and the rate of consumption of hydrogen is given by: \( \frac{R_{NH_3}}{2}=\frac{R_{H_2}}{3} \)
03

Solve for the rate of production of ammonia

Now, let's solve for the rate of production of ammonia, \(R_{NH_3}\), in terms of the rate of consumption of hydrogen, \(R_{H_2}\). We can rearrange the equation in Step 2 to find the rate of production of ammonia: \(R_{NH_3}=2\times\frac{R_{H_2}}{3}\)
04

Write the final relationship

The relationship between the rate of production of ammonia and the rate of consumption of hydrogen in the Haber process is given by: \[ R_{NH_3}=\frac{2}{3} R_{H_2} \] This equation states that the rate of production of ammonia is 2/3 times the rate of consumption of hydrogen.

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Most popular questions from this chapter

Consider the hypothetical reaction \(\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(g) \longrightarrow\) \(2 \mathrm{AB}(g)\), where the rate law is: $$ -\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{~A}_{2}\right]\left[\mathrm{B}_{2}\right] $$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is \(2.45 \times 10^{-4} \mathrm{~L} / \mathrm{mol}\). \(\mathrm{s}\), and at \(508^{\circ} \mathrm{C}\) the rate constant is \(0.891 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\). What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C}\) ?

Assuming that the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(12.7\) is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\) or \(\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?\) How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(12.7 ?\)

Would the slope of \(a \ln (k)\) versus \(1 / T\) plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the \(\ln (k)\) versus \(1 / T\) plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.

A reaction of the form \(\mathrm{aA} \longrightarrow\) Products gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3}\). Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M\), calculate the time (in seconds) required for the reaction to reach \(22.9 \%\) completion.

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0\), and \(40.0 \mathrm{~min}\) for an experiment in which \([\mathrm{A}]_{0}=0.10 M .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. \(80.0 \mathrm{~min}\) b. \(30.0 \mathrm{~min}\)
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