Chapter 7: Problem 61
The equilibrium constant for the reaction \(2 \mathrm{X}(\mathrm{g})+\) \(\mathrm{Y}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Z}(\mathrm{g})\) is \(2.25\) litre \(\mathrm{mol}^{-1} .\) What would be the concentration of \(\mathrm{Y}\) at cquilibrium with \(2.0\) moles of \(\mathrm{X}\) and \(3.0\) molcs of \(\mathrm{Z}\) in \(1-\mathrm{L}\) vessel? (1) \(1.0 \mathrm{M}\) (2) \(2.25 \mathrm{M}\) (3) \(2.0 \mathrm{M}\) (4) \(4.0 \mathrm{M}\)
Short Answer
Expert verified
[\text{Y}] = 1.0 \text{ M}
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
equilibrium constant
The equilibrium constant, denoted as \( K_c \), is a crucial concept in chemical equilibrium. It represents the ratio of the concentration of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced chemical equation. For a given reaction, if the value of \( K_c \) is known, we can predict the extent of the reaction and the concentrations of various species at equilibrium. In our example, the reaction is given by:
\[ 2 \text{X(g)} + \text{Y(g)} \rightleftharpoons 2 \text{Z(g)} \]
The equilibrium constant expression for this reaction is:
\[ K_c = \frac{[\text{Z}]^2}{[\text{X}]^2[\text{Y}]} \]
Here, \[ K_c = 2.25 \text{ litre mol}^{-1} \]. This means that at equilibrium, the ratio of the products to the reactants will always equal 2.25, regardless of the initial concentrations, assuming the temperature remains constant.
\[ 2 \text{X(g)} + \text{Y(g)} \rightleftharpoons 2 \text{Z(g)} \]
The equilibrium constant expression for this reaction is:
\[ K_c = \frac{[\text{Z}]^2}{[\text{X}]^2[\text{Y}]} \]
Here, \[ K_c = 2.25 \text{ litre mol}^{-1} \]. This means that at equilibrium, the ratio of the products to the reactants will always equal 2.25, regardless of the initial concentrations, assuming the temperature remains constant.
concentration calculations
Concentration calculations are essential when dealing with chemical equilibrium. These calculations involve determining the molarity (moles per liter) of reactants and products at equilibrium. In the given exercise, we start with initial concentrations of \[ [\text{X}] \] and \[ [\text{Z}] \]. Given that \[ [\text{X}] \] is 2.0 M (moles per liter) and \[ [\text{Z}] \] is 3.0 M, we use these values in the equilibrium expression.
Substituting these known values into the equilibrium constant expression:
\[ 2.25 = \frac{(3.0)^2}{(2.0)^2[\text{Y}]} \]
Next, we simplify the expression:
\[ 2.25 = \frac{9.0}{4.0[\text{Y}]} \]
By multiplying both sides by 4.0[\text{Y}], we isolate [\text{Y}] on one side:
\[ 2.25 \times 4.0[\text{Y}] = 9.0 \]
This simplifies to:
\[ 9.0[\text{Y}] = 9.0 \]
Finally, dividing both sides by 9.0 gives us the concentration of \[ \text{Y} \] at equilibrium:
\[ [\text{Y}] = 1.0 \text{ M} \]
Substituting these known values into the equilibrium constant expression:
\[ 2.25 = \frac{(3.0)^2}{(2.0)^2[\text{Y}]} \]
Next, we simplify the expression:
\[ 2.25 = \frac{9.0}{4.0[\text{Y}]} \]
By multiplying both sides by 4.0[\text{Y}], we isolate [\text{Y}] on one side:
\[ 2.25 \times 4.0[\text{Y}] = 9.0 \]
This simplifies to:
\[ 9.0[\text{Y}] = 9.0 \]
Finally, dividing both sides by 9.0 gives us the concentration of \[ \text{Y} \] at equilibrium:
\[ [\text{Y}] = 1.0 \text{ M} \]
reaction quotient
The reaction quotient, denoted as \[ Q \], helps us understand the direction in which a reaction will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant but with the initial concentrations instead of equilibrium concentrations.
Mathematically, for the reaction:
\[ 2 \text{X(g)} + \text{Y(g)} \rightleftharpoons 2 \text{Z(g)} \]
the reaction quotient expression is:
\[ Q = \frac{[\text{Z}]_0^2}{[\text{X}]_0^2[\text{Y}]_0} \]
Here, \[ [\text{Z}]_0 \], \[ [\text{X}]_0 \], and \[ [\text{Y}]_0 \] are the initial concentrations. If \[ Q < K_c \], the reaction will shift to the right, forming more products. Conversely, if \[ Q > K_c \], the reaction will shift to the left, forming more reactants. If \[ Q = K_c \], the system is at equilibrium. Understanding \[ Q \] helps predict how concentrations will change until the system reaches equilibrium.
In our case, the initial concentrations were given as \[ [\text{X}]_0 = 2.0 \text{ M} \] and \[ [\text{Z}]_0 = 3.0 \text{ M} \] for a 1-L vessel. We used the equilibrium constant expression directly since the system was assumed to be at equilibrium, but the concept of the reaction quotient would help in determining the direction if these were initial concentrations.
Mathematically, for the reaction:
\[ 2 \text{X(g)} + \text{Y(g)} \rightleftharpoons 2 \text{Z(g)} \]
the reaction quotient expression is:
\[ Q = \frac{[\text{Z}]_0^2}{[\text{X}]_0^2[\text{Y}]_0} \]
Here, \[ [\text{Z}]_0 \], \[ [\text{X}]_0 \], and \[ [\text{Y}]_0 \] are the initial concentrations. If \[ Q < K_c \], the reaction will shift to the right, forming more products. Conversely, if \[ Q > K_c \], the reaction will shift to the left, forming more reactants. If \[ Q = K_c \], the system is at equilibrium. Understanding \[ Q \] helps predict how concentrations will change until the system reaches equilibrium.
In our case, the initial concentrations were given as \[ [\text{X}]_0 = 2.0 \text{ M} \] and \[ [\text{Z}]_0 = 3.0 \text{ M} \] for a 1-L vessel. We used the equilibrium constant expression directly since the system was assumed to be at equilibrium, but the concept of the reaction quotient would help in determining the direction if these were initial concentrations.
