Question

the reaction
$2nobr(g)\
ightarrow 2no(g) + br_2(g)$
is a second order reaction with a rate constant of $0.80\\ m^{-1}s^{-1}$ at $11\\ ^\circ c$. if the initial concentration of nobr is $0.0440\\ m$, the concentration of nobr after 10.0 seconds is

0.0275 m
0.0350 m
0.0325 m
0.0400 m
0.0300 m

Explanation:

Step1: Recall 2nd-order reaction formula

$$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$$
Where:

• $[A]_t$ = concentration at time $t$,
• $[A]_0$ = initial concentration,
• $k$ = rate constant,
• $t$ = time.

Step2: Plug in given values

Given: $[A]_0=0.0440\ \text{M}$, $k=0.80\ \text{M}^{-1}\text{s}^{-1}$, $t=10.0\ \text{s}$
$$\frac{1}{[A]_t} = \frac{1}{0.0440} + (0.80)(10.0)$$

Step3: Calculate each term

$$\frac{1}{0.0440} \approx 22.7273, \quad (0.80)(10.0)=8.0$$
$$\frac{1}{[A]_t} = 22.7273 + 8.0 = 30.7273$$

Step4: Solve for $[A]_t$

$$[A]_t = \frac{1}{30.7273} \approx 0.0325\ \text{M}$$

Answer:

0.0325 M