The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g)$$ is 4.2 at \(1650^{\circ} \mathrm{C}\). Initially \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.80 \mathrm{~mol}\) \(\mathrm{CO}_{2}\) are injected into a 5.0 - \(\mathrm{L}\) flask. Calculate the concentration of each species at equilibrium.

Short Answer

Expert verified
To find the short answer, calculate the value of x from step 3 and substitute it in step 4 equation to get the equilibrium concentrations of [H_2], [CO_2], [H_2O] and [CO].

Step by step solution

01

Write the Reaction and Equilibrium Expression

The reaction is \(H_{2}(g) + CO_{2}(g) \rightleftharpoons H_{2}O(g) + CO(g)\). The equilibrium expression for this reaction is \(K_c = \frac{[H_{2}O][CO]}{[H_{2}][CO_{2}]}\).
02

Set up the ICE Table

Here, the initial number of moles for \(H_2\) and \(CO_2\) is 0.8 mol each. As for \(H_2O\) and \(CO\), initially there's no product in our system, so the initial moles are 0. The equilibrium concentrations can be calculated by dividing the number of moles by the volume of the flask which is 5.0L. Since initially there were no products, the initial concentrations are:[H_2] = [CO_2] = 0.8 mol / 5.0L = 0.16 M. For \(H_2O\) and \(CO\), [H_2O] = [CO] = 0 M.During the reaction, we lose x amount of \(H_2\) and \(CO_2\) and gain the same x amount of \(H_2O\) and \(CO\). Hence, the change is represented as:[H_2] = [CO_2] = -x [H_2O] = [CO] = +x The equilibrium concentrations then become:[H_2] = [CO_2] = 0.16 - x [H_2O] = [CO] = x
03

Substitute in the Equilibrium Expression and Solve for x

Substitute the obtained equilibrium concentrations into the equation for \(K_c\) and solve for x.Putting these values on the equilibrium constant expression, we get:4.2 = (x * x) / ((0.16 - x) * (0.16 - x)) Solve the above equation to find out the value of x.
04

Calculate the Equilibrium Concentrations

Substitute the value of x back into the equilibrium concentrations to get the concentrations of each species at equilibrium:[H_2] = [CO_2] = 0.16 - x[H_2O] = [CO] = x

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is a pivotal concept in chemistry, essential for grasping how reactions proceed and reach a state of balance. It occurs when a system's forward reaction (reactants turning into products) and reverse reaction (products re-forming into reactants) proceed at the same rate. As a result, the concentrations of all reactants and products remain constant over time, meaning they do not show any noticeable change despite the continuous occurrence of both reactions.

In the provided exercise, chemical equilibrium is achieved in the reaction involving hydrogen, carbon dioxide, water, and carbon monoxide gases. When a system reaches equilibrium, it's crucial to note that the process doesn't halt; both the forward and reverse reactions are ongoing, just at a rate that doesn't change the overall quantities of reactants and products.

To measure this dynamic balance, chemists use the equilibrium constant, denoted as K_c in the case of concentrations, which provides a ratio reflecting the relationship between the concentrations of products and reactants at equilibrium.
ICE Table
One effective tool for solving equilibrium problems is the ICE Table, which stands for Initial, Change, and Equilibrium. This is a practical method for organizing what's known and what needs to be determined about the reactants and products in a chemical reaction. It's like a ledger that keeps track of where the reaction starts, how it changes, and where it eventually balances out.

In our exercise, the ICE Table is used to track the initial moles of hydrogen and carbon dioxide, both set at 0.8 mole. Since no products were present initially, their moles are counted as zero. As the reaction proceeds, we document the change for each substance: losing 'x' moles of reactants and gaining 'x' moles of products. This simple accounting allows us to express the final equilibrium concentrations in terms of 'x', which is our unknown value representing the extent of the reaction at equilibrium.

Using an ICE Table effectively simplifies complicated equilibrium calculations by providing a systematic approach to determine concentration changes throughout the reaction.
Equilibrium Expressions
Equilibrium expressions are mathematical representations that relate the concentrations of the products and reactants at equilibrium for a chemical reaction. These expressions are based on the law of mass action which states that for a reversible reaction at equilibrium and at a constant temperature, a certain ratio of the concentrations of reactants and products is constant, known as the equilibrium constant (K).

In this particular exercise, the equilibrium expression is given by the formula:
\[K_c = \frac{[H_{2}O][CO]}{[H_{2}][CO_{2}]}\]
The square brackets indicate the molar concentrations of each species at equilibrium. The expression for the equilibrium constant, K_c, demonstrates that its value is determined by the concentration of the products raised to the power of their coefficients and divided by the concentration of the reactants raised to the power of their coefficients as shown in a balanced chemical equation.

To find the equilibrium concentrations, we substitute the expressions from the ICE Table into the equilibrium constant formula and solve for 'x'. Once 'x' is known, we can then determine the concentrations of all species at equilibrium. Properly setting up and calculating equilibrium expressions is critical in predicting the behavior of a chemical system under various conditions.

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Most popular questions from this chapter

List four factors that can shift the position of an equilibrium. Only one of these factors can alter the value of the equilibrium constant. Which one is it?

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$\mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g)$$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C},\) respectively. \()\)

Baking soda (sodium bicarbonate) undergoes thermal decomposition as follows: $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

Consider the equilibrium system \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). Sketch the changes in the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) over time for the following situations: (a) Initially only A is present. (b) Initially only B is present. (c) Initially both A and B are present (with A in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

Consider the following equilibrium system involving \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (sulfuryl dichloride): $$\mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)$$ Predict how the equilibrium position would change if (a) \(\mathrm{Cl}_{2}\) gas were added to the system; (b) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) were removed from the system; (c) \(\mathrm{SO}_{2}\) were removed from the system. The temperature remains constant.

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