1. Consider the reaction 2H3PO4 yields P2O5+3H2O Using the information in the fo
ID: 988611 • Letter: 1
Question
1. Consider the reaction
2H3PO4 yields P2O5+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)..............P2O5 (M)
0.........................0
10.......................2.50*10^-3
20......................5.50*10^-3
30 ................... 7.30*10^-3
40.....................8.50*10^-3
50 .....................9.10*10^-3
Rate of formation of P2O5 = ?
what is the average rate of decomposition of H3PO4 between 10.0 and 40.0 s?
2. Consider the reaction:
5Br-(aq) + BrO-3(aq) + 6H+(aq) ---> 3Br2(aq) + 3H20(l)
The average rate of consumption of Br- is 1.06 x 10^-4 M/s over the first two minutes. What is the average rate of formation of Br2 during the same time interval?
Rate of formation of Br2 = ?
Explanation / Answer
1) rate = - 1/2 d/dt [H3PO4] = d/dt [P2O5] = 1/3 d/dt [H2O] rate avg = x2-x1/dt2-dt1
rate of formation of [P2O5] between 10 and 40 = - 1/2 [x40 - x10]/ t40-t10
=- 1/2 [8.50 *10-3 - 2.50*10-3]/40-10 = 6 *10-3/ 30 = 2 *10-2 M sec
as - 1/2 d/dt [H3PO4] = d/dt [P2O5] putting value we get
rate of decomposition of [H3PO4] =- - 2 * rate of formation of [P2O5] = 2 * 2 *10-2 M sec = 4 *10-2 M sec
2) rate = - 1/5 d[BR-] / dt ( rate of decomposition of[BR-]) = 1/3 d[BR2]/dt ( rate of formation of [BR2)
rate of formation of Br2 = 3/5 rate of decompositon of Br-
= 3/5* 1.06 x 10^-4 M/s = 6.36 * 10 -5 M sec-1
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.