Problem 108
Question
Consider the following reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ The rate of this reaction in terms of \(\mathrm{N}_{2}\) at \(\mathrm{T}\) is \(-\mathrm{d}\left[\mathrm{N}_{2}\right] /\) \(\mathrm{dt}=0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) What is the value of \(-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}\) (in units of \(\left.\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) at the same temperature? (a) \(0.02\) (b) 50 (c) \(0.06\) (d) \(0.04\)
Step-by-Step Solution
Verified Answer
The value is 0.06 mol L⁻¹ s⁻¹ (option c).
1Step 1: Understand the Rate Expression
The reaction is balanced as follows: \[ \text{N}_{2}(\mathrm{g}) + 3\text{H}_{2}(\mathrm{g}) \longrightarrow 2\text{NH}_{3}(\mathrm{g}) \] The rate expression involving reactants tells us how fast the concentration of the reactants is decreasing. For \(\text{N}_{2}\), the rate of disappearance is given as \(-\frac{d[\text{N}_{2}]}{dt} = 0.02 \text{ mol L}^{-1} \text{ s}^{-1}\). We need to find \(-\frac{d[\text{H}_{2}]}{dt}\).
2Step 2: Use the Stoichiometry of the Reaction
According to the balanced chemical equation, 1 mol of \(\text{N}_{2}\) reacts with 3 mol of \(\text{H}_{2}\). This means that \(\text{H}_{2}\) is used up at a rate that is three times faster than \(\text{N}_{2}\). Therefore, the rate of disappearance of \(\text{H}_{2}\) can be expressed as \(-\frac{d[\text{H}_{2}]}{dt} = 3 \times (-\frac{d[\text{N}_{2}]}{dt})\).
3Step 3: Calculate \(-\frac{d[H_2]}{dt}\)
Substituting the given rate of disappearance for \(\text{N}_{2}\): \[ -\frac{d[\text{H}_{2}]}{dt} = 3 \times 0.02 \text{ mol L}^{-1} \text{ s}^{-1}\] Calculate the value: \[ -\frac{d[\text{H}_{2}]}{dt} = 0.06 \text{ mol L}^{-1} \text{ s}^{-1}\]
4Step 4: Select the Correct Answer
The calculated rate of disappearance of \(\text{H}_{2}\) is \(0.06 \text{ mol L}^{-1} \text{ s}^{-1}\). Looking at the provided options, the correct choice is \(c)\) \(0.06\).
Key Concepts
Reaction RateChemical StoichiometryRate of Reaction
Reaction Rate
The reaction rate is a measure of how quickly a chemical reaction occurs. It is typically expressed in terms of how fast the concentration of a reactant decreases or how fast the concentration of a product increases, over time. For example, in our given reaction involving nitrogen \( \text{N}_2 \\) and hydrogen \( \text{H}_2 \\), the rate is shown as a change in concentration over a change in time. \Understanding the reaction rate helps chemists to control and optimize reactions for industrial processes. It can vary based on different factors such as temperature, concentration of reactants, and presence of catalysts. \
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- High reaction rates mean a reaction proceeds quickly to completion, which is beneficial in manufacturing. \
- Low reaction rates can lead to inefficiencies or incomplete reactions, undesirable in many settings. \
Chemical Stoichiometry
Chemical stoichiometry gives us the quantitative relationship between reactants and products in a balanced chemical equation. It provides the ratio in which chemicals combine to form a product. In our reaction: \[ \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}) \] stoichiometry tells us that: \
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- 1 mole of nitrogen gas \(\text{N}_2\\) reacts with 3 moles of hydrogen gas \(\text{H}_2\\). \
- These reactants produce 2 moles of ammonia \(\text{NH}_3\\). \
Rate of Reaction
The rate of reaction specifies how fast reactants are converted into products. In our example, the rate has been given for nitrogen \(\text{N}_2\\) as \(-\frac{d[\text{N}_2]}{dt} = 0.02\ \text{mol L}^{-1} \text{s}^{-1}\). This tells us how quickly \(\text{N}_2\\) is being consumed. \To find the rate of reaction for \(\text{H}_2\\), which is consumed at a different rate due to stoichiometry, we apply the stoichiometric relationship discussed earlier:
- For every mole of \(\text{N}_2\\) consumed, 3 moles of \(\text{H}_2\\) are consumed. \
Other exercises in this chapter
Problem 106
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