*Need answer to PART D based on previous answers Now consider the following reac

Question

*Need answer to PART D based on previous answers

Now consider the following reaction and data:

H2+2IBr?2HBr+I2

Part B

Part complete

What is the average rate of formation of I2?

0.025 Ms

Part C

Part complete

Based on your answer to Part B, what is the average rate of formation of HBr?

0.050 Ms

According to the chemical equation, there will be twice as many moles of HBr formed as moles of I2. Therefore, the molar concentration of HBr is increasing at twice the rate of the I2 concentration.

Part D

Based on your answer to Part B or C,

What is the average rate of change of H2?

Remember that reactant concentrations decrease over time.

Explanation / Answer

H2   +    2IBr   <------------------------> 2HBr   +   I2

rate = - d[H2] / dt = - 1/2 d [IBr] / dt = 1/ 2 d[HBr] /dt = d[I2] /dt

d[I2] /dt   = (1.67 - 1.42) / (15-5) = 0.025 M / s

part D)

- d[H2] / dt = d[I2] /dt

average rate change of H2 = - 0.025 M / s

minus symbol indicates decreasing the concenration of reactants

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