The gas-phase decomposition of NOBr has a k = 0.810 M-1 s-1 at 10°C. NOBr (g) → NO (g) + ½ Br2 (g) rate = k[NOBr]2 We start with 0.00200 M NOBr in a flask at 10°C. (a) How many minutes does it take to use up 0.00050 M of this NOBr? (b) If we start with 0.000500 M NOBr, what concentration of NOBr with remain after 5.00 minutes of reaction? (c) How much of the NO is produced after 5 min when the reaction started with 0.000500 M NOBr?

a) 2.57 min

b) 6.075x10⁻⁵ M

c) 4.3925x10⁻⁴ M

Explanation:

a) The rate determines how the concentration of NOBr decays at each second (M/s), thus, if the initial concetrations is [NOBr] = 0.00200, thus the rate is:

rate = 0.810 * (0.002) ²

rate = 3.24x10⁻⁶ M/s

So, at each second, 3.24x10⁻⁶ M is used, thus, the time to use up 0.00050 M is:

t = 0.00050/3.24x10⁻⁶

t = 154.32 s

t = 2.57 min

b) Know the rate will be calculated with [NOBr] = 0.00050 M, so:

rate = 0.810 * (0.00050) ²

rate = 2.025x10⁻⁷ M/s

So, after 5 minutes = 300s, the concentration decays to:

2.025x10⁻⁷ M/s * 300s = 6.075x10⁻⁵ M

c) If remains 6.075x10⁻⁵ M of NOBr, than it was consumed:

0.00050 - 6.075x10⁻⁵ = 4.3925x10⁻⁴ M

The stoichiometry of the reaction is 1 mole of NOBr to 1 mole of NO, so it is formed 4.3925x10⁻⁴ M of NO.