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Chapter 11: Problem 8

Ammonia is produced by the reaction between nitrogen and hydro- gen gases. (a) Write a balanced equation using smallest whole-number coefficients for the reaction. (b) Write an expression for the rate of reaction in terms of \(\Delta\left[\mathrm{NH}_{3}\right]\). (c) The concentration of ammonia increases from \(0.257 \mathrm{M}\) to \(0.815 \mathrm{M}\) in \(15.0 \mathrm{~min} .\) Calculate the average rate of reaction over this time interval.

Short Answer

Expert verified
Answer: The average rate of reaction is \(6.2 \times 10^{-4} \frac{\mathrm{M}}{\mathrm{s}}\).

Step by step solution

01

Write the balanced equation for the reaction

To produce ammonia (NH3), nitrogen (N2) reacts with hydrogen (H2) in the given proportion: N2 + 3H2 -> 2NH3
02

Write an expression for the rate of reaction

The rate of reaction is the change in the concentration of ammonia over the change in time. We can represent the rate of reaction (r) as follows: \[r = \frac{\Delta\left[\mathrm{NH}_{3}\right]}{\Delta t}\]
03

Calculate the initial and final concentrations of ammonia

The initial concentration of ammonia is 0.257 M and the final concentration is 0.815 M.
04

Calculate the change in concentration of ammonia

We need to find the difference between the initial and final concentrations: \[\Delta\left[\mathrm{NH}_{3}\right] = \text{Final concentration} - \text{Initial concentration}\] \[\Delta\left[\mathrm{NH}_{3}\right] = 0.815\mathrm{M} - 0.257\mathrm{M}\] \[\Delta\left[\mathrm{NH}_{3}\right] = 0.558\mathrm{M}\]
05

Convert the time interval to seconds

The time interval given is 15.0 minutes. Converting it to seconds: \[15.0\text{ minutes} \times \frac{60\text{ seconds}}{1\text{ minute}} = 900\text{ seconds}\]
06

Calculate the average rate of reaction

Now we can plug the change in concentration of ammonia and the time interval into the rate of reaction formula: \[r = \frac{0.558\mathrm{M}}{900\mathrm{s}}\] \[r = 6.2 \times 10^{-4} \frac{\mathrm{M}}{\mathrm{s}}\] The average rate of reaction over this time interval is \(6.2 \times 10^{-4} \frac{\mathrm{M}}{\mathrm{s}}\).

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Most popular questions from this chapter

For the zero-order decomposition of HI on a gold surface $$\mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{I}_{2}(g)$$ it takes \(16.0 \mathrm{~s}\) for the pressure of HI to drop from \(1.00\) atm to \(0.200 \mathrm{~atm}\). (a) What is the rate constant for the reaction? (b) How long will it take for the pressure to drop from \(0.150 \mathrm{~atm}\) to \(0.0432\) atm? (c) What is the half-life of \(\mathrm{HI}\) at a pressure of \(0.500 \mathrm{~atm}\) ?

Nitrosyl bromide (NOBr) decomposes to nitrogen oxide and bromine. Use the following data to determine the order of the decomposition reaction of nitrosyl bromide. $$\begin{array}{cccccc}\hline \text { Time (s) } & 0 & 6 & 12 & 18 & 24 \\ \text { [NOBr] } & 0.0286 & 0.0253 & 0.0229 & 0.0208 & 0.0190 \\\\\hline\end{array}$$

A catalyst lowers the activation energy of a reaction from \(215 \mathrm{~kJ}\) to \(206 \mathrm{~kJ} .\) By what factor would you expect the reaction-rate constant to increase at \(25^{\circ} \mathrm{C} ?\) Assume that the frequency factors \((\mathrm{A})\) are the same for both reactions. (Hint: Use the formula \(\ln k=\ln \mathrm{A}-E_{2} / R T .\) )

Write the rate expression for each of the following elementary steps: (a) \(\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}\) (b) \(\mathrm{I}_{2} \longrightarrow 2 \mathrm{I}\) (c) \(\mathrm{NO}+\mathrm{O}_{2} \longrightarrow \mathrm{NO}_{3}\)

The first-order rate constant for the decomposition of a certain hormone in water at \(25^{\circ} \mathrm{C}\) is \(3.42 \times 10^{-4}\) day \(^{-1}\). (a) If a \(0.0200 \mathrm{M}\) solution of the hormone is stored at \(25^{\circ} \mathrm{C}\) for two months, what will its concentration be at the end of that period? (b) How long will it take for the concentration of the solution to drop from \(0.0200 M\) to \(0.00350 \mathrm{M} ?\) (c) What is the half-life of the hormone?
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