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Sia

青州's Question
Chemistry
Posted 2 months ago

Consider the following reaction in a batch reactor at constant volume:
NO(g)+H2(g)N2(g)+H2O(g)
\mathrm{NO}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)

(a) If hydrogen is in excess, how does the formation rate of nitrogen and water relate to the reaction rate of nitric oxide?
(b) Determine the composition of the reactor if the fractional conversion is 0.8 and the initial composition was 10 molL1NO10 \mathrm{~mol} \mathrm{L-1} \mathrm{NO} and 20 molL1H220 \mathrm{~mol} \mathrm{L-1} \mathrm{H}_{2}
(c) If the reaction is first order in NO\mathrm{NO} and second order in H2\mathrm{H}_{2}. Write the rate law for the reaction and find the units for the rate constant if the reaction progress is measured in concentration ( molL1)\mathrm{mol} \mathrm{L-1)}.
(d) If the reaction progress is measured in partial pressure (Pa)(\mathrm{Pa}), write the rate law and find the units for the rate constant.
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Answer from Sia
Posted 2 months ago
Solution
1
Stoichiometry of the reaction: The balanced chemical equation is 2NO(g)+2H2(g)N2(g)+2H2O(g)2\mathrm{NO}(g) + 2\mathrm{H}_2(g) \rightarrow \mathrm{N}_2(g) + 2\mathrm{H}_2\mathrm{O}(g). This indicates that 2 moles of NO react with 2 moles of H₂ to form 1 mole of N₂ and 2 moles of H₂O
2
Relation of formation rates: Since hydrogen is in excess, it does not limit the reaction rate. The formation rate of N₂ and H₂O is directly related to the consumption rate of NO by the stoichiometry of the reaction
a Answer
The formation rate of N₂ and H₂O is half the reaction rate of NO when H₂ is in excess.
Key Concept
Stoichiometry and reaction rates
Explanation
The formation rates of products in a chemical reaction are directly related to the consumption rates of reactants according to the stoichiometry of the balanced chemical equation.
1
Initial moles of NO and H₂: Initially, there are 10molL110 \, \mathrm{mol \, L^{-1}} of NO and 20molL120 \, \mathrm{mol \, L^{-1}} of H₂
2
Moles of NO reacted: With a fractional conversion of 0.8, 0.8×10molL1=8molL10.8 \times 10 \, \mathrm{mol \, L^{-1}} = 8 \, \mathrm{mol \, L^{-1}} of NO have reacted
3
Moles of NO remaining: 10molL18molL1=2molL110 \, \mathrm{mol \, L^{-1}} - 8 \, \mathrm{mol \, L^{-1}} = 2 \, \mathrm{mol \, L^{-1}} of NO remain
4
Moles of H₂ remaining: Since H₂ is in excess and the stoichiometry is 1:1, we assume there is still excess H₂ left
5
Moles of N₂ formed: 8molL18 \, \mathrm{mol \, L^{-1}} of NO reacted to form 4molL14 \, \mathrm{mol \, L^{-1}} of N₂
6
Moles of H₂O formed: Similarly, 8molL18 \, \mathrm{mol \, L^{-1}} of NO reacted to form 8molL18 \, \mathrm{mol \, L^{-1}} of H₂O
b Answer
After a fractional conversion of 0.8, the composition is 2molL12 \, \mathrm{mol \, L^{-1}} NO, excess H₂, 4molL14 \, \mathrm{mol \, L^{-1}} N₂, and 8molL18 \, \mathrm{mol \, L^{-1}} H₂O.
Key Concept
Fractional conversion and stoichiometry
Explanation
The composition of the reactor after a certain fractional conversion can be determined by using the initial amounts of reactants and the stoichiometry of the balanced chemical equation.
1
Rate law expression: The rate law for the reaction, given it is first order in NO and second order in H₂, is rate=k[NO][H2]2rate = k[\mathrm{NO}][\mathrm{H}_2]^2
2
Units for rate constant: The overall order of the reaction is 1+2=31 + 2 = 3. The units of the rate constant kk are mol1Ls1\mathrm{mol^{-1} \, L \, s^{-1}} for a third-order reaction when concentration is used
c Answer
The rate law is rate=k[NO][H2]2rate = k[\mathrm{NO}][\mathrm{H}_2]^2 with units for kk being mol1L2s1\mathrm{mol^{-1} \, L^2 \, s^{-1}}.
Key Concept
Rate law and reaction order
Explanation
The rate law expresses the rate of a chemical reaction in terms of the concentration of reactants and the rate constant, whose units depend on the overall reaction order.
1
Rate law with partial pressures: The rate law, when expressed in terms of partial pressures, is rate=k[PNO][PH2]2rate = k[P_{\mathrm{NO}}][P_{\mathrm{H}_2}]^2
2
Units for rate constant with partial pressures: The units for the rate constant kk when using partial pressures are Pa2s1\mathrm{Pa^{-2} \, s^{-1}} for a third-order reaction
d Answer
The rate law is rate=k[PNO][PH2]2rate = k[P_{\mathrm{NO}}][P_{\mathrm{H}_2}]^2 with units for kk being Pa2s1\mathrm{Pa^{-2} \, s^{-1}}.
Key Concept
Rate law with partial pressures
Explanation
The rate law can also be expressed in terms of partial pressures, and the units of the rate constant will change accordingly to reflect the different units of measurement.

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