Time, min |
NH3, M |
0 |
6 |
1 |
3 |
2 |
1.5 |
3 |
0.75 |
4 |
0.375 |
5 |
0.1875 |
Show your work.
Solution:
From the table,
Part A) When, t = 0 min, then [NH3] = 6 M
At ,t = 1 min, [NH3] = 3 M
Thus,
Initial rate = - dX/dt = - (3 M - 6 M) / 1 - 0 min
Initial rate = 3.0 M /min
Part B)
Average rate = - dX/dt
dX = final concentration - initial concentration
dX = 0.1875 M - 6.0 M = -5.8125 M
dt = 5 -0 = 5 min
Therefore,
Average rate = - (- 5.8125) M / 5 min
= 1.1625 M/min
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