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The rate of disappearance of HBr in the gas phase reaction 2HBR (g) → H2 (g) + Br2 (g) is 0.310 M/sec at 150°C. The relative rate of appearance of Br, is M/s.

The rate of disappearance of HBr in the gas phase reaction 2HBR (g) → H2 (g) + Br2 (g) is 0.310 M/sec at 150°C. The relative rate of appearance of Br, is M/s.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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### Understanding Reaction Rates in Chemical Kinetics The rate of disappearance of HBr in the gas phase reaction is given by the equation: \[2HBr \, (g) \rightarrow H_2 \, (g) + Br_2 \, (g) \] According to the problem statement, the rate of disappearance of HBr at a temperature of 150°C is \(0.310 \, \text{M/sec}\). The task then is to determine the relative rate of appearance of \(Br_2 \). To find the relative rate of the appearance of \(Br_2\), we use the stoichiometry of the reaction. In the balanced chemical equation, you can see that: \[2 \, \text{mol of} \, HBr \, \text{produce 1 mol of} \, Br_2 \] This implies that the rate at which \(Br_2\) appears is half the rate at which \(HBr\) disappears: \[\text{Rate of appearance of } Br_2 = \frac{1}{2} \times \text{(Rate of disappearance of } HBr) \] Given that the rate of disappearance of HBr is \(0.310 \, \text{M/sec}\): \[\text{Rate of appearance of } Br_2 = \frac{1}{2} \times 0.310 \, \text{M/sec} = 0.155 \, \text{M/sec}\] Thus, the relative rate of appearance of \(Br_2\) is \(0.155 \, \text{M/sec}\). This fundamental understanding of reaction rates helps in various applications within chemical kinetics and is crucial for predicting the behavior of reactions over time.
Transcribed Image Text:### Understanding Reaction Rates in Chemical Kinetics The rate of disappearance of HBr in the gas phase reaction is given by the equation: \[2HBr \, (g) \rightarrow H_2 \, (g) + Br_2 \, (g) \] According to the problem statement, the rate of disappearance of HBr at a temperature of 150°C is \(0.310 \, \text{M/sec}\). The task then is to determine the relative rate of appearance of \(Br_2 \). To find the relative rate of the appearance of \(Br_2\), we use the stoichiometry of the reaction. In the balanced chemical equation, you can see that: \[2 \, \text{mol of} \, HBr \, \text{produce 1 mol of} \, Br_2 \] This implies that the rate at which \(Br_2\) appears is half the rate at which \(HBr\) disappears: \[\text{Rate of appearance of } Br_2 = \frac{1}{2} \times \text{(Rate of disappearance of } HBr) \] Given that the rate of disappearance of HBr is \(0.310 \, \text{M/sec}\): \[\text{Rate of appearance of } Br_2 = \frac{1}{2} \times 0.310 \, \text{M/sec} = 0.155 \, \text{M/sec}\] Thus, the relative rate of appearance of \(Br_2\) is \(0.155 \, \text{M/sec}\). This fundamental understanding of reaction rates helps in various applications within chemical kinetics and is crucial for predicting the behavior of reactions over time.
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