Question

a chemical engineer is studying the rate of this reaction. nh₄oh(aq)→nh₃(aq)+h₂o(aq) she fills a reaction vessel with nh₄oh and measures its concentration as the reaction proceeds. heres a graph of her data: use this graph to answer the following questions: what is the half - life of the reaction? round your answer to 2 significant digits. suppose the rate of the reaction is known to be first order in nh₄oh. calculate the value of the rate constant k. round your answer to 2 significant digits. also be sure you include the correct unit symbol. predict the concentration of nh₄oh in the engineers reaction vessel after 200 seconds have passed. assume no other reaction is important, and continue to assume the rate is first order in nh₄oh. round your answer to 2 significant digits.

Explanation:

Step1: Determine half - life from graph

The half - life ($t_{1/2}$) of a reaction is the time it takes for the concentration of the reactant to decrease to half of its initial value. From the graph, if we assume the initial concentration of $\text{NH}_4\text{OH}$ is around $1.8\ M$ (from the y - axis at $t = 0$), half of this value is $0.9\ M$. Reading the time on the x - axis corresponding to a concentration of $0.9\ M$, we find $t_{1/2}\approx20\ s$.

Step2: Calculate rate constant for first - order reaction

For a first - order reaction, the relationship between the half - life and the rate constant ($k$) is given by the formula $k=\frac{\ln2}{t_{1/2}}$. Substituting $t_{1/2}=20\ s$ into the formula, we get $k = \frac{\ln2}{20\ s}\approx0.035\ s^{-1}$.

Step3: Predict concentration using first - order integrated rate law

The integrated rate law for a first - order reaction is $\ln\frac{[\text{A}]_t}{[\text{A}]_0}=-kt$. Let $[\text{A}]_0 = 1.8\ M$ (initial concentration), $k = 0.035\ s^{-1}$, and $t = 200\ s$. First, calculate $-kt=- 0.035\ s^{-1}\times200\ s=-7$. Then, $\ln\frac{[\text{A}]_t}{[\text{A}]_0}=-7$. Exponentiating both sides gives $\frac{[\text{A}]_t}{[\text{A}]_0}=e^{-7}$. So, $[\text{A}]_t=[\text{A}]_0\times e^{-7}=1.8\ M\times e^{-7}\approx0.0012\ M\approx1.2\times 10^{-3}\ M$.

Answer:

$t_{1/2}=20\ s$
$k = 0.035\ s^{-1}$
$[\text{NH}_4\text{OH}]=1.2\times 10^{-3}\ M$