Ammonium cyanate, NHNCO, rearranges in water to give urea, (NH2)2 CO: NH4 NCO(aq) ⟶(NH2)2 CO(aq)

step 1 Number 61 is a moderately lengthy problem. The first thing that you're going to have to do is gr

Ammonium cyanate, NHNCO, rearranges in water to give urea,...

Ammonium cyanate, NH_NCO, rearranges in water to give urea, $\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}:$ $$\mathrm{NH}_{4} \mathrm{NCO}(\mathrm{aq}) \longrightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(\mathrm{aq})$$ $$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\(\min )\end{array} & \begin{array}{l}{\left[\mathrm{NH}_{4} \mathrm{NCO}\right]} \\(\mathrm{mol} / \mathrm{L})\end{array} \\\hline 0 & 0.458 \\4.50 \times 10^{1} & 0.370 \\1.07 \times 10^{2} & 0.292 \\2.30 \times 10^{2} & 0.212 \\6.00 \times 10^{2} & 0.114 \\\hline\end{array}$$ Using the data in the table: (a) Decide whether the reaction is first order or second order. (b) Calculate $k$ for this reaction. (c) Calculate the half-life of ammonium cyanate under these conditions. (d) Calculate the concentration of $\mathrm{NH}_{4} \mathrm{NCO}$ after $12.0 \mathrm{h}$.

Ammonium cyanate, NH_NCO, rearranges in water to give urea, $\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}:$
$$\mathrm{NH}_{4} \mathrm{NCO}(\mathrm{aq}) \longrightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(\mathrm{aq})$$
$$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\(\min )\end{array} & \begin{array}{l}{\left[\mathrm{NH}_{4} \mathrm{NCO}\right]} \\(\mathrm{mol} / \mathrm{L})\end{array} \\\hline 0 & 0.458 \\4.50 \times 10^{1} & 0.370 \\1.07 \times 10^{2} & 0.292 \\2.30 \times 10^{2} & 0.212 \\6.00 \times 10^{2} & 0.114 \\\hline\end{array}$$
Using the data in the table:
(a) Decide whether the reaction is first order or second order.
(b) Calculate $k$ for this reaction.
(c) Calculate the half-life of ammonium cyanate under these conditions.
(d) Calculate the concentration of $\mathrm{NH}_{4} \mathrm{NCO}$ after $12.0 \mathrm{h}$.

Key Concepts

Half-Life

The half-life of a reaction is defined as the time required for the concentration of a reactant to decrease to half of its initial value. In first order reactions, the half-life is constant and independent of the initial concentration, given by t?/? = ln2/k. In contrast, for second order reactions, the half-life depends on the initial concentration, making it variable. Understanding half-life is essential for predicting how long a reaction will take to reach a certain extent of completion.

Rate Constant (k)

The rate constant is a fundamental parameter in kinetic studies that quantifies the speed of a chemical reaction. Its value is obtained from the slope of the linear plot in the appropriate integrated rate law and indicates the proportionality between the reaction rate and the concentration(s) raised to the power of their order. The units of the rate constant vary depending on whether the reaction is first order (s?¹) or second order (M?¹·s?¹).

Concentration Extrapolation Using Kinetic Equations

By applying the integrated rate law and knowing the rate constant, one can predict the concentration of reactants at any given time. This extrapolation is a common application in chemical kinetics to forecast the progress of a reaction under specified conditions. It uses the determined relationship between concentration and time to calculate future concentrations, which is particularly useful for planning and optimizing chemical processes.

Reaction Order

Reaction order describes how the rate of a chemical reaction depends on the concentration of its reactants. Determining the reaction order involves analyzing how changes in concentration affect the rate, and this information is inferred by comparing experimental data to the predictions made by integrated rate laws. The reaction can be first order, where the rate is proportional to the concentration of one reactant, or second order, where the rate is proportional to the square of the concentration or the product of concentrations of two reactants.

Integrated Rate Laws

Integrated rate laws provide mathematical expressions that relate the concentrations of reactants to time during the course of a reaction. For a first order reaction, the integrated rate law is expressed as ln[A] = -kt + ln[A]?, leading to a linear relationship when plotting the natural logarithm of concentration versus time. For a second order reaction, the law is typically 1/[A] = kt + 1/[A]?. These equations are crucial for determining the reaction order and the rate constant from experimental data.

Transcript

00:01 Number 61 is a moderately lengthy problem.

00:04 The first thing that you're going to have to do is graph the data that's provided.

00:09 This ends up being a graphical rate law question, where after graphing the data, you can determine the order of the reaction, you can determine the rate constant, and then knowing the rate constant, you can answer the other parts of this question.

00:26 There's really no other way to determine the order of the reaction.

00:30 Unless you graph the data.

00:33 So the first thing that i did then was graphed the data.

00:37 I went into excel.

00:39 I took my time values that were given to me.

00:44 I took the concentration values that were given to me.

00:47 I then took the natural log of these concentration values and the one over the concentration values in order to get a two graphs that were representative of the natural log of the concentration as a function of time and one over the concentration as a function of time.

01:13 Although it didn't ask for the third graph, i had the data, so i went ahead and i plotted just the concentration as a function of time.

01:22 And we see that when we plot the concentration as a function of time, there's definitely a curve.

01:28 When we see that we plot the natural log of the concentration as a function of time, there's a curve.

01:34 But when we plot the concentration as a function of time, we do get a straight line.

01:40 We wouldn't get a straight line when plotting the one over the concentration as a function of time unless it were anything but second order...

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