Use the following table to determine the average rate during the period 800 to 1200 seconds for the decomposition of H2O2.

2 H2O2(l) → 2 H2O(l) + O2(g)

Time (sec) (H2O2)M
0 2.32
400 1.72
800 1.30
1200 0.98
1600 0.73

a] 1.2 x 10-3M/sec
b] 1.6 x 10-3 M/sec
c] 2.4 x 10-3M/sec
d] 8.2 x 10-4M/sec
e] 8.0 x 10-4M/sec

the spacing came out weird so just to clarify, the times are 0, 400, 800, 1200, and 1600 secs.

8.0x10-4

wrong 8.0x10-4

To determine the average rate during the period 800 to 1200 seconds for the decomposition of H2O2, we need to calculate the change in concentration of H2O2 over that time interval and divide it by the time interval.

Given:
Time (sec): 800 1200
(H2O2)M: 1.30 0.98

The change in concentration of H2O2 over 800 to 1200 seconds is:
Change in [H2O2] = Final [H2O2] - Initial [H2O2]
= 0.98 M - 1.30 M
= -0.32 M

Note: The negative sign indicates that the concentration of H2O2 decreases over time, as it decomposes.

The time interval is:
Time interval = Final time - Initial time
= 1200 sec - 800 sec
= 400 sec

Now, we can calculate the average rate using the formula:

Average Rate = Change in [H2O2] / Time interval

Average Rate = (-0.32 M) / (400 sec)

Calculating this gives us approximately -8.0 x 10-4 M/sec.

Therefore, the answer is e] 8.0 x 10-4M/sec. The average rate during the period 800 to 1200 seconds for the decomposition of H2O2 is approximately 8.0 x 10-4M/sec.