One way to calculate the value of a reaction's equilibrium constant is to perform the reaction, let it come to equilibrium, measure the concentration of all the reactants and products, and then plug those concentrations into the equilibrium constant expression and calculate its value. (a) A student performs the reaction \(\mathrm{A}(a q)+2 \mathrm{~B}(a q) \rightleftarrows \mathrm{C}(a q)\) starting with \(2.0 \mathrm{M} \mathrm{A}\) and \(4.0 \mathrm{M} \mathrm{B}\). He finds that at equilibrium, the concentrations of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) are \(0.020 \mathrm{M}, 0.040 \mathrm{M}\), and \(1.98\) \(\mathrm{M}\), respectively. What is the value of this reaction's equilibrium constant (write your answer using scientific notation)? (b) Next, the student repeats the experiment, but this time he starts with \(3.0 \mathrm{M} \mathrm{A}\) and 5.0 M B. What value will he get for \(K_{\text {eq }}\) when he measures the equilibrium concentrations and plugs them into the equilibrium constant expression? (Hint: Think about why equilibrium constant is called a constant.)

Step by step solution

01

(a) Identifying the reaction and equilibrium concentrations

In this scenario, the given chemical reaction is: $$\mathrm{A}(a q)+2 \mathrm{~B}(a q) \rightleftarrows \mathrm{C}(a q)$$ The student starts with the initial concentrations: - \(\mathrm{A}\): \(2.0\) M - \(\mathrm{B}\): \(4.0\) M At equilibrium, the concentrations are: - \(\mathrm{A}\): \(0.020\) M - \(\mathrm{B}\): \(0.040\) M - \(\mathrm{C}\): \(1.98\) M

02

(a) Defining the equilibrium constant expression

According to the given chemical equation, the equilibrium constant expression for this reaction is: $$K_\mathrm{eq}=\frac{[\mathrm{C}]}{[\mathrm{A}]\times[\mathrm{B}]^2}$$ Now, we will plug in the equilibrium concentrations of A, B and C.

03

(a) Calculating the equilibrium constant

Plugging the equilibrium concentrations into the equilibrium constant expression: $$K_\mathrm{eq}=\frac{1.98}{(0.020)(0.040)^2}$$ Calculate the value of \(K_\mathrm{eq}\): $$K_\mathrm{eq}=4.125 \times 10^4$$ The value of the reaction's equilibrium constant is \(4.125 \times 10^4\).

04

(b) Impact of different initial concentrations on equilibrium constant

Now, the student performs the reaction with different initial concentrations: - \(\mathrm{A}\): \(3.0\) M - \(\mathrm{B}\): \(5.0\) M As the name suggests, the equilibrium constant is constant for a given reaction at a specific temperature, regardless of the initial concentrations. Therefore, even if the student uses different initial concentrations of A and B, the equilibrium constant (\(K_\mathrm{eq}\)) will still be the same. So, when the student measures the equilibrium concentrations for this experiment and plugs them into the equilibrium constant expression, he will get the same value for \(K_\mathrm{eq}\): $$K_\mathrm{eq} = 4.125 \times 10^4$$

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