Problem 16 - Chemistry and Chemical Reactivity

Question

Data for the reaction $$\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{HPO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ are provided in the table. $$\begin{array}{llll}\hline & & & \text { Initial Rate } \\\\\text { Experiment } & {\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right](\mathrm{M})} &{\left[\mathrm{OH}^{-}\right](\mathrm{M})} & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\\\\hline 1 & 0.0030 & 0.00040 & 0.0020 \\\2 & 0.0030 & 0.00080 & 0.0080 \\\3 & 0.0090 & 0.00040 & 0.0060 \\\4 & ? & 0.00033 & 0.0020 \\\\\hline\end{array}$$ (a) What is the rate law for this reaction? (b) What is the value of \(k ?\) (c) What is the concentration of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) in experiment 4?

Step-by-Step Solution

Answer

(a) Rate = k [H2PO4^-] [OH^-]^2. (b) k = 4.167 × 10^6 mol^-2 L^2 min^-1. (c) [H2PO4^-] = 0.00492 M.

Step 1: Determine the Rate Law Form

The rate law for the reaction is generally given by the form:\[\text{Rate} = k [\text{H}_2\text{PO}_4^-]^m [\text{OH}^-]^n\]where \(m\) and \(n\) are the orders with respect to \(\text{H}_2\text{PO}_4^-\) and \(\text{OH}^-\), respectively. These need to be determined from the experimental data.

Step 2: Determine the Order with Respect to OH^-

Compare experiments 1 and 2 where \([\text{H}_2\text{PO}_4^-]\) is constant while \([\text{OH}^-]\) doubles:- Experiment 1: \(\text{Rate} = 0.0020\), \([\text{OH}^-] = 0.00040\)- Experiment 2: \(\text{Rate} = 0.0080\), \([\text{OH}^-] = 0.00080\)The rate increased by factor of 4 which suggests that the reaction is second order in \([\text{OH}^-]\). Hence, \(n = 2\).

Step 3: Determine the Order with Respect to H2PO4^-

Compare experiments 1 and 3 where \([\text{OH}^-]\) is constant:- Experiment 1: \(\text{Rate} = 0.0020\), \([\text{H}_2\text{PO}_4^-] = 0.0030\)- Experiment 3: \(\text{Rate} = 0.0060\), \([\text{H}_2\text{PO}_4^-] = 0.0090\)The rate increased by a factor of 3, which suggests that the reaction is first order in \([\text{H}_2\text{PO}_4^-]\). Hence, \(m = 1\).

Step 4: Write the Rate Law

From Steps 2 and 3, the rate law is:\[\text{Rate} = k [\text{H}_2\text{PO}_4^-]^1 [\text{OH}^-]^2\]So, the rate law is: \(\text{Rate} = k [\text{H}_2\text{PO}_4^-] [\text{OH}^-]^2\).

Step 5: Calculate the Rate Constant k

Using data from Experiment 1:\[0.0020 = k (0.0030)^1 (0.00040)^2\]\[k = \frac{0.0020}{0.0030 \times (0.00040)^2} = \frac{0.0020}{0.0030 \times 0.00000016} = 4.1667 \times 10^6 \text{ mol}^{-2} \text{ L}^2 \text{ min}^{-1}\]Thus, \( k \approx 4.167 \times 10^6 \text{ mol}^{-2} \text{ L}^2 \text{ min}^{-1} \).

Step 6: Calculate Concentration of H2PO4^- in Experiment 4

In Experiment 4, we use the solved rate law and known values:\[0.0020 = 4.167 \times 10^6 [\text{H}_2\text{PO}_4^-] (0.00033)^2\]\[[\text{H}_2\text{PO}_4^-] = \frac{0.0020}{4.167 \times 10^6 \times (0.00033)^2} \]\[= \frac{0.0020}{4.167 \times 10^6 \times 0.0000001089} \approx 0.00492 \text{ M}\]Thus, \([\text{H}_2\text{PO}_4^-] \approx 0.00492 \text{ M}\).

Key Concepts

Rate Law

The rate law is an essential part of understanding reaction kinetics. It expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. The general form of a rate law for a reaction like \[ ext{A} + ext{B} \rightarrow ext{Products} \]can be expressed as:\[ \text{Rate} = k [ ext{A}]^m [ ext{B}]^n \]Here, \( k \) is the rate constant, and \( m \) and \( n \) represent the reaction order with respect to each reactant. In this case, our reaction involves \( \text{H}_2\text{PO}_4^- \) and \( \text{OH}^- \). By examining experimental data, these variables—\( m \) and \( n \)—are determined, which define the reaction order.

Reaction Order

Reaction order indicates how the rate of reaction is influenced by the concentration of reactants. To find the reaction order in this example, we compare various experiments:

• With constant \([\text{H}_2\text{PO}_4^-]\), doubling \([\text{OH}^-]\) results in the reaction rate increasing fourfold. This tells us that the reaction is second order in \([\text{OH}^-]\) (\(n = 2\)).
• Keeping \([\text{OH}^-]\) constant while tripling \([\text{H}_2\text{PO}_4^-]\) leads to a tripling of the rate. This suggests a first order reaction with respect to \([\text{H}_2\text{PO}_4^-]\) (\(m = 1\)).

Understanding these changes is crucial, as reaction order dictates the rate law.

Rate Constant

The rate constant, \( k \), embodies the speed of a reaction given the concentration of reactants. It is unique to each reaction and varies with temperature but remains unchanged with concentration. To find the rate constant from experimental data, use the derived rate law:\[ \text{Rate} = k [\text{H}_2\text{PO}_4^-]^1 [\text{OH}^-]^2 \]Plug in the known values of rate and concentrations:\[ 0.0020 = k (0.0030)^1 (0.00040)^2 \]Solving for \( k \), we find:\[ k \approx 4.167 \times 10^6 \text{\, mol}^{-2} \text{\, L}^2 \text{\, min}^{-1} \]This constant provides crucial insight into the reaction dynamics.

Concentration Calculation

Often, one might need to find unknown reactant concentrations in a reaction using the rate equation and known values. For experiment 4, where \([\text{H}_2\text{PO}_4^-]\) is unknown, we use:\[ 0.0020 = 4.167 \times 10^6 [\text{H}_2\text{PO}_4^-] (0.00033)^2 \]By re-arranging and solving for \([\text{H}_2\text{PO}_4^-]\), we find:\[ [\text{H}_2\text{PO}_4^-] \approx 0.00492 \text{\, M} \]This calculation ensures that we can anticipate the concentration of reactants given the specifics of the rate and the effect of other components. Knowing how to manage these calculations is vital for dynamic control and prediction in reaction kinetics.