Question

under certain conditions the rate of the reaction is zero - order in nitrogen monoxide with a rate constant of 0.0032 m s⁻¹. 2n₂o(g)→2n₂(g)+o₂(g) suppose a 3.0 l flask is charged under these conditions with 350 mmol of nitrogen monoxide. how much is left 6.0 s later? you may assume no other reaction is important. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

Explanation:

Step1: Identify the rate - law and relevant values

The rate law for a zero - order reaction is $rate = k$, where $k = 0.0032\ mol\cdot L^{-1}\cdot s^{-1}$. The initial amount of $N_2O$ is $n_0=3.0\ mol$ (since $V = 1.0\ L$ and initial concentration $[N_2O]_0 = 3.0\ mol/L$), and the time $t = 6.0\ s$.

Step2: Use the integrated rate law for zero - order reaction

The integrated rate law for a zero - order reaction is $[A]_t=[A]_0−kt$. In terms of moles, $n_t=n_0−ktV$. Substitute $n_0 = 3.0\ mol$, $k = 0.0032\ mol\cdot L^{-1}\cdot s^{-1}$, $t = 6.0\ s$, and $V = 1.0\ L$ into the equation.
$n_t=3.0\ mol-(0.0032\ mol\cdot L^{-1}\cdot s^{-1}\times6.0\ s\times1.0\ L)$.

Step3: Calculate the amount of $N_2O$ left

First, calculate the amount of $N_2O$ reacted: $0.0032\ mol\cdot L^{-1}\cdot s^{-1}\times6.0\ s\times1.0\ L=0.0192\ mol$. Then, find the amount of $N_2O$ left: $n_t = 3.0\ mol - 0.0192\ mol=2.9808\ mol$. Rounding to the correct number of significant figures (3 significant figures as 3.0 has 2 and 6.0 has 2, and in subtraction we consider the least number of decimal places), $n_t\approx2.98\ mol$.

Answer:

$2.98\ mol$