Question

What is the average rate of production of Br2 over the first 10 sec for 2 NOBr(g)  Br2(g) + 2 NO(g)? Time (s) [NOBr] (M) 0.00s 0.0100M 2.00s 0.0071M 4.00s 0.005M5 6.00s 0.0045M 8.00s 0.0038M 10.00s 0.0033M

Answer 1

The balanced reaction is

2NOBr(g) ----------------------- Br2(g) + 2NO(g)

Average rate of NO over first 10 seconds = [[NO](t=10s) - [NO](t=0)]/(10-0) = (0.0033-0)/10 = 0.00033M/s

Now as per the balanced reaction, one mole of Br2 is formed for every 2 moles of NO formed

Hence, average rate of production of Br2 over first 10 seconds = 1/2 * average rate of NO over first 10 seconds = 1/2 * 0.00033 = 0.000165 M/s