Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Learning Goal: To calculate average and relative reaction rates. You can measure

ID: 1022718 • Letter: L

Question

Learning Goal: To calculate average and relative reaction rates. You can measure the rate of a reaction, just like you can measure the speed a jogger runs. While a jogger would be reported to run a specific number of miles in an hour, miles/hour, a reaction is reported to form product or consume reagent in molar concentration per second, M/s. Reaction rate can be defined either as the increase in the concentration of a product per unit time or as the decrease of the concentration of a reactant per unit time. By definition, reaction rate is a positive quantity. In the reaction X2Y, for example, Y is being produced twice as fast as X is consumed and thus rate of X=12(rate of Y) Each rate can be expressed as the change in concentration over the change in time, t: [X]t=12([Y]t) Consider the reaction 2H3PO4P2O5+3H2O Using the information in the following table, calculate the average rate of formation of P2O5 between 10.0 and 40.0 s. Time (s) 0 10.0 20.0 30.0 40.0 50.0 [P2O5] (M) 0 1.70×103 4.70×103 6.50×103 7.70×103 8.30×103 Part A Rate of formation of P2O5 = 2.00×104 Ms it is done Part B Determine the average rate of decomposition of H3PO4 between 10.0 and 40.0 s. Express your answer with the appropriate units. Part C Consider the reaction 5Br(aq)+BrO3(aq)+6H+(aq)3Br2(aq)+3H2O(l) The average rate of consumption of Br is 1.86×104 M/s over the first two minutes. What is the average rate of formation of Br2 during the same time interval? Express your answer with the appropriate units.

Explanation / Answer

2H3PO4P2O5+3H2O

Time (s)..............P2O5 (M)

0.........................0

10.......................1.7*10^-3

20......................4.70*10^-3

30 ................... 6.50*10^-3

40.....................7.70*10^-3

50 .....................8.3*10^-3

Rate of formation of P2O5 =2.00×104 M/s

PartA.

Average rate of formation of P2O5 between 10.0 and 40.0 s. = P2O5 / t

                                                                                                     = (7.70*10^-3 -1.7*10^-3) M/ (40-10)s

                                                                                                     = 6*10^-3M/30s = 2.00×104 M/s

Part B

Reaction equation shows that every two molecules of H3PO4 disappear one molecule of P2O5 is formed.

The average rate of decomposition of H3PO4 between 10.0 and 40.0 s

Rate of reaction = rate of formation of P2O5

Rate of reaction is also =1/2 rate of decomposition of H3PO4

.: Rate of decomposition of H3PO4= 2* rate of formation of P2O5

.: Rate of decomposition of H3PO4= 2*2.00×104 M/s

                                                                                  = 4.00×104 M/s

PartC.

5Br1- (aq) + BrO-3(aq)+ 6H+ (aq) = 3Br2(aq) + 3 H2O(l)

Reaction equation shows that 5 Br- disappear 3 molecules of Br2 is formed

The average rate of consumption of Br is 1.86×104 M/s over the first two minutes

The average rate of formation of Br2 during the same time interval

                                                                                  = The average rate of consumption of Br x 3/5

                                                                                  =1.86×104 M/s x 3/5

                                                                                  = 1.116×104 M/s

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote