The kinetics of the decomposition of phosphine at 950 K 4 PH3(g) ⟶P4(g)+6 H2(g) was studied by i

step 1 This question is similar to question number nine. In order to determine the rate constant of the

The kinetics of the decomposition of phosphine at 950 K 4...

The kinetics of the decomposition of phosphine at $950 \mathrm{K}$ $$4 \mathrm{PH}_{3}(\mathrm{g}) \longrightarrow \mathrm{P}_{4}(\mathrm{g})+6 \mathrm{H}_{2}(\mathrm{g})$$ was studied by injecting $\mathrm{PH}_{3}(\mathrm{g})$ into a reaction vessel and measuring the total pressure at constant volume. $$\begin{array}{lc} \hline P_{\text {total }}(\text { Torr }) & \text { Time (s) } \\ \hline 100 & 0 \\ 150 & 40 \\ 167 & 80 \\ 172 & 120 \\ \hline \end{array}$$ What is the rate constant of this reaction?

The kinetics of the decomposition of phosphine at $950 \mathrm{K}$
$$4 \mathrm{PH}_{3}(\mathrm{g}) \longrightarrow \mathrm{P}_{4}(\mathrm{g})+6 \mathrm{H}_{2}(\mathrm{g})$$
was studied by injecting $\mathrm{PH}_{3}(\mathrm{g})$ into a reaction vessel and measuring the total pressure at constant volume.
$$\begin{array}{lc}
\hline P_{\text {total }}(\text { Torr }) & \text { Time (s) } \\
\hline 100 & 0 \\
150 & 40 \\
167 & 80 \\
172 & 120 \\
\hline
\end{array}$$
What is the rate constant of this reaction?

Key Concepts

Constant Volume Conditions

When a reaction is carried out in a vessel of constant volume, changes in the total pressure are directly related to changes in the amount of gaseous species present. This relationship allows one to use pressure measurements to track the progress of a reaction and, in turn, to determine kinetic parameters such as the rate constant.

Reaction Kinetics

This concept involves studying the rate at which chemical reactions proceed and the factors that influence these rates. It includes measuring how concentrations (or pressures, in the case of gases) change with time and using these changes to infer details about the mechanism and order of the reaction.

Rate Constant

The rate constant is a numerical value that quantifies the speed of a reaction under given conditions. It appears in the rate law, which relates the rate of a reaction to the concentrations of the reactants, and is specific to a reaction at a given temperature.

Integrated Rate Law

An integrated rate law is an expression that relates the concentration (or pressure) of a reactant or product to time, based on the order of the reaction. It is used to linearize experimental data so that the rate constant can be determined from the slope of the resulting plot.

Reaction Stoichiometry in Gaseous Systems

Reaction stoichiometry relates the quantitative relationships between reactants and products. In gaseous reactions, changes in the number of moles of gases affect measurable properties like the total pressure under constant volume, and these relationships must be considered when analyzing reaction kinetics.

Transcript

00:01 This question is similar to question number nine.

00:05 In order to determine the rate constant of the reaction, if we are given concentration data or pressure data as a function of time, is to plot the concentration or the pressure as a function of time and then plot one over the concentration or one over the concentration or pressure as a function of time and the natural log of pressure or concentration as a function of time for the single reactant.

00:35 However, for this particular problem, we do not know the pressure of the single reactant as a function of time.

00:42 We are provided with the total pressure of everything in the mixture, the reactants or the reactant and the products as a function of time.

00:53 So we need to calculate just the pressure of the reactant ph3 as a function of time so that we can prepare the graphs as i described them.

01:06 The total pressure is going to be a function of the pressure of ph3 because it's a gas and the two products, p4 and h2.

01:18 So the total pressure will be equal to the pressure of ph3 plus the pressure of p4 plus the pressure of p4 plus the pressure of of h2.

01:29 At 40 seconds, we know that the total pressure is 150 tor.

01:36 So that will be equal to the pressure of ph3, which will be 100 tor minus the change, because it had a pressure of 100 tor at times zero.

01:49 Because there's a 4 coefficient in front of ph 3, we can express the change as minus 4x.

01:58 Then we will create p4, with a coefficient of 1 in front of it, its pressure will simply be x, one -fourth the change, and then for hydrogen with a coefficient of 6 in front of it, its pressure will be 6x.

02:15 This now gives us a single equation that allows us to solve for x, which in this case is 16 .67 tor.

02:24 So now the pressure of just ph 3, which is what we need, will be 100 minus 4x7.

02:31 4 times 16 .67...

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