at a certain temperature the rate of this reaction is first order in nh₄oh with a rate constant of 3.73 s⁻¹…

at a certain temperature the rate of this reaction is first order in nh₄oh with a rate constant of 3.73 s⁻¹. nh₄oh(aq)→nh₃(aq)+h₂o(aq) suppose a vessel contains nh₄oh at a concentration of 1.43 m. calculate the concentration of nh₄oh in the vessel 0.310 seconds later. you may assume no other reaction is important. round your answer to 2 significant digits.

at a certain temperature the rate of this reaction is first order in nh₄oh with a rate constant of 3.73 s⁻¹. nh₄oh(aq)→nh₃(aq)+h₂o(aq) suppose a vessel contains nh₄oh at a concentration of 1.43 m. calculate the concentration of nh₄oh in the vessel 0.310 seconds later. you may assume no other reaction is important. round your answer to 2 significant digits.

Answer

Explanation:

Step1: Recall first - order reaction formula

For a first - order reaction, the integrated rate law is $\ln\left(\frac{[A]_t}{[A]_0}\right)=-kt$, where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time $t$, $k$ is the rate constant, and $t$ is the time. We are given $[A]_0 = 1.43M$, $k = 3.73s^{-1}$, and $t=0.310s$.

Step2: Rearrange the formula to solve for $[A]_t$

Starting from $\ln\left(\frac{[A]_t}{[A]_0}\right)=-kt$, we can exponentiate both sides: $\frac{[A]_t}{[A]_0}=e^{-kt}$. Then $[A]_t=[A]_0e^{-kt}$.

Step3: Substitute the given values

$[A]_t = 1.43M\times e^{-(3.73s^{-1}\times0.310s)}$. First, calculate the exponent: $-(3.73\times0.310)= - 1.1563$. Then, find $e^{-1.1563}\approx0.315$. So, $[A]_t=1.43M\times0.315 = 0.45045M$.

Step4: Round to 2 significant digits

Rounding $0.45045M$ to 2 significant digits gives $0.45M$.

Answer:

$0.45M$