Question

The reaction 2NoBr(g) ---> 2NO(g)+Br2(g) is a second order reaction with a rate constant of 0.80M^-1 s^-1 at 11 C. If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 12 seconds is _________.

Answer 1

The concentration of NOBr after 12 seconds is 0.0408 M.

Step-by-step explanation:

The given reaction is 2NOBr(g) → 2NO(g) + Br2(g), which is a second-order reaction with a rate constant of 0.80 M-1 s-1 at 11°C. To find the concentration of NOBr after 12 seconds, we can use the integrated rate law for a second-order reaction:

[NOBr]t = [NOBr]0 / (1 + kt), where [NOBr]t is the concentration after time t, [NOBr]0 is the initial concentration, k is the rate constant, and t is the time. Plugging in the values, we get:

[NOBr]12 = 0.0440 M / (1 + (0.80 M-1 s-1) * (12 s)) = 0.0408 M

Answer 2

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Step-by-step explanation: