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Chapter 12: Problem 26

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 (NH3) and the rate of consumption of hydrogen (H2) in the Haber process is given by: \(\text{rate of consumption of H}_{2} = \frac{3}{2} \times \text{rate of production of NH}_{3}\). This means that for every 1 mole of ammonia produced, 3/2 moles of hydrogen are consumed.

Step by step solution

01

Identify the balanced chemical equation

First, let's identify the balanced chemical equation for the Haber process for the production of ammonia: \[ \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \]
02

Determine the stoichiometric coefficients

We can see the stoichiometric coefficients for each component in the reaction: - For nitrogen (N2), the coefficient is 1. - For hydrogen (H2), the coefficient is 3. - For ammonia (NH3), the coefficient is 2. These coefficients represent the molar ratio of the reactants and products in the reaction.
03

Establish the relationship between the rate of consumption and production in the reaction

The rate of a chemical reaction can be expressed in terms of the rate of change of concentration (in moles per unit time) of the reactants and products. In this case, we are looking for the relationship between the rate of consumption of hydrogen (H2) and the rate of production of ammonia (NH3). Using the stoichiometric coefficients, we can write the relationship as follows: \[ \frac{\text{rate of consumption of H}_{2}}{3} = -\frac{1}{3}\frac{d[\mathrm{H}_{2}]}{dt} = \frac{\text{rate of production of NH}_{3}}{2} \]
04

Solve for the rate of consumption of hydrogen

Now, we will solve for the rate of consumption of hydrogen (H2) in terms of the rate of production of ammonia (NH3): \[ \text{rate of consumption of H}_{2} = 3\left(\frac{\text{rate of production of NH}_{3}}{2}\right) \]
05

Write the final relationship

The final relationship between the rate of production of ammonia and the rate of consumption of hydrogen is given by: \[ \text{rate of consumption of H}_{2} = \frac{3}{2} \times \text{rate of production of NH}_{3} \] This means that for every 1 mole of ammonia produced, 3/2 moles of hydrogen are consumed in the Haber process.

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Most popular questions from this chapter

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text {Rate} =-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\) \([\mathrm{B}]_{0}=3.0 M,\) and \([\mathrm{C}]_{0}=2.0 M .\) The reaction is started, and after 8.0 seconds, the concentration of \(\mathrm{A}\) is \(3.8 \times 10^{-3} \mathrm{M}\) a. Calculate the value of k for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after 13.0 seconds. d. Calculate the concentration of C after 13.0 seconds.

Draw a rough sketch of the energy profile for each of the following cases: a. \(\Delta E=+10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=25 \mathrm{kJ} / \mathrm{mol}\) b. \(\Delta E=-10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\) c. \(\Delta E=-50 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\)

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 \(\mathrm{kJ} / \mathrm{mol} .\) In the presence of a catalyst at \(37^{\circ} \mathrm{C}\) the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

The activation energy for some reaction $$ \mathrm{X}_{2}(g)+\mathrm{Y}_{2}(g) \longrightarrow 2 \mathrm{XY}(g) $$ is 167 \(\mathrm{kJ} / \mathrm{mol}\) , and \(\Delta E\) for the reaction is \(+28 \mathrm{kJ} / \mathrm{mol}\) . What is the activation energy for the decomposition of XY?

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of \(\Delta \mathrm{E}\) . c. The half-life for a reaction, aA \(\longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0} .\) d. The half-life for a second-order reaction increases as the reaction proceeds.
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