## Related questions with answers

Question

The rate constant for the second-order reaction

$2 \mathrm{NOBr}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{~g})$

is $0.80 \mathrm{M}^{-1} \mathrm{~s}^{-1}$ at $10^{\circ} \mathrm{C}$.

Starting with a concentration of $0.086 \mathrm{M}$, calculate the concentration of $\mathrm{NOBr}$ after 22 s.

Solution

VerifiedAnswered 1 year ago

Answered 1 year ago

Step 1

1 of 5In this exercise, we need to calculate the concentration of NOBr after 22 seconds that the below-shown reaction has been ongoing:

$\mathrm{2NOBr(g)\rightarrow 2NO(g)+Br_2(g)}$

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