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The rate law for the decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) is $$ \text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right] $$ It takes 120. s for \(1.00 M \mathrm{PH}_{3}\) to decrease to \(0.250 \mathrm{M}\). How much time is required for \(2.00 M \mathrm{PH}_{3}\) to decrease to a concentration of \(0.350 M ?\)

Short Answer

Expert verified
The time required for 2.00 M PH3 to decrease to a concentration of 0.350 M is 132 seconds.

Step by step solution

01

Write the rate law expression.

The given rate law expression for the decomposition of phosphine (PH3) is: $$ \text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right] $$
02

Find the rate constant k using the given information.

To find the rate constant k, we can use the given initial concentration, final concentration, and time for the first scenario: Initial concentration (\([\mathrm{PH}_{3}]_{i}\)) = 1.00 M Final concentration (\([\mathrm{PH}_{3}]_{f}\)) = 0.250 M Change in concentration (\(\Delta [\mathrm{PH}_{3}]\)) = \((1.00 - 0.250) \, \mathrm{M} = 0.750 \, \mathrm{M}\) Time (Δt) = 120 s Plugging this information into the rate law expression, we get: $$ \text { Rate }=-\frac{0.750 \, \mathrm{M}}{120 \, \text{s}}=k\left(1.00 \, \mathrm{M}\right) $$ Now, we can solve for k: $$ k=\frac{-\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t\left[\mathrm{PH}_{3}\right]_i}=\frac{-0.750 \, \mathrm{M}}{(120 \, \text{s})(1.00 \, \mathrm{M})} = -0.00625\,\frac{\text {M}}{\text{s}} $$ Since k should be positive, we'll take the absolute value of the obtained constant: $$ k=0.00625\,\frac{\text {M}}{\text{s}} $$
03

Equation to solve for new time using new concentrations.

Now, we have k and we can use it to find the time required for the second scenario, where the initial concentration is 2.00 M and the final concentration is 0.350 M. We can set up the rate law equation with this new information: Initial concentration (\([\mathrm{PH}_{3}]_{i}\)) = 2.00 M Final concentration (\([\mathrm{PH}_{3}]_{f}\)) = 0.350 M Change in concentration (\(\Delta [\mathrm{PH}_{3}]\)) = \((2.00 - 0.350) \, \mathrm{M} = 1.650 \, \mathrm{M}\) Rate constant (k) = 0.00625 M/s $$ \text { Rate }=-\frac{\Delta\left[\mathrm{PH}_{3}\right]}{\Delta t}=k\left[\mathrm{PH}_{3}\right]_i $$ $$ -\frac{1.650 \, \mathrm{M}}{\Delta t}=0.00625\,\frac{\text {M}}{\text{s}}(2.00 \, \mathrm{M}) $$
04

Solve for the new time.

To calculate the new time, we can rearrange the equation and solve for Δt: $$ \Delta t=\frac{-\Delta\left[\mathrm{PH}_{3}\right]}{k\left[\mathrm{PH}_{3}\right]_i}=\frac{-1.650 \, \mathrm{M}}{(0.00625\, \frac{\text {M}}{\text{s}})(2.00 \, \mathrm{M})} = -132 \, \text{s} $$ Since the time should be positive, we'll take the absolute value of the obtained time: $$ \Delta t=132 \, \text{s} $$ So, it takes 132 seconds for 2.00 M PH3 to decrease to a concentration of 0.350 M.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Rate
Understanding how quickly a chemical reaction occurs is essential for many fields of science and engineering. This is where the concept of the 'reaction rate' comes into play. In simple terms, the reaction rate indicates how fast the concentration of a reactant decreases or the concentration of a product increases over time. The formula
\[\text{Reaction rate} = -\frac{\Delta[\mathrm{A}]}{\Delta t}\]
represents the rate at which a reactant A is consumed in a chemical reaction. The negative sign indicates the concentration of the reactant is decreasing. In the exercise, this was showcased by finding how fast phosphine (\(\mathrm{PH}_3\)) decomposes.
To ensure students grasp this concept fully, incorporating visual aids like graphs showing concentration versus time can be beneficial. It's important to note that reaction rate can be affected by various factors such as temperature, catalysts, and concentration of reactants, which could form part of a more advanced lesson.
Rate Law
Diving deeper into kinetics, the 'rate law' expresses the relationship between the reaction rate and the concentration of reactants. This mathematical equation reflects how the rate depends on the concentration of one or more reactants. In the given exercise, the rate law for the decomposition of phosphine is written as
\[\text{Rate} = k[\mathrm{PH}_3]\]
where \(k\) is the rate constant, and [\(\mathrm{PH}_3\)] represents the concentration of phosphine. Students should learn that the rate law is specific to each reaction and must be determined experimentally; it cannot be deduced from the chemical equation alone. To clarify the concept, providing examples of different types of rate laws, like zero-order, first-order, or second-order can illustrate how the rate varies with concentration.
Rate Constant
At the heart of the rate law lies the 'rate constant', symbolized by \(k\). This coefficient is a measure of the intrinsic reactivity of the reaction, and it remains constant at a given temperature. The rate constant essentially provides a link between the concentration of the reactants and the speed at which they react. In quantitative terms, it's a proportionality factor in the rate law equation, as seen in the problem's solution where
\[k = 0.00625 \, \frac{\text{M}}{\text{s}}\]
was calculated. Despite its name, the rate constant can be influenced by environmental conditions like temperature, with its magnitude providing insight into the reaction's speed. A larger \(k\) means a faster reaction at the same concentration levels. Simple animations or simulations showing how altering \(k\) influences the rate can be very instructive at driving this point home.

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Most popular questions from this chapter

One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{aligned} &\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \\ &\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ Overall reaction \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{a}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) $$ is \(14.0 \mathrm{~kJ} . E_{a}\) for the same reaction when catalyzed is \(11.9 \mathrm{~kJ}\). What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction.

One reason suggested for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below: The activation energy for such a process is \(210 \mathrm{~kJ} / \mathrm{mol}\), which is less than either the \(\mathrm{Si}-\) Si or the \(\mathrm{Si}-\mathrm{H}\) bond energy. Why would a similar mechanism not be expected to play a very important role in the decomposition of long chains of carbon atoms as seen in organic compounds?

Define stability from both a kinetic and thermodynamic perspective. Give examples to show the differences in these concepts.

Assuming that the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(12.7\) is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\) or \(\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?\) How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(12.7 ?\)

Would the slope of \(a \ln (k)\) versus \(1 / T\) plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the \(\ln (k)\) versus \(1 / T\) plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.

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