For the synthesis of ammonia $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ the equilibrium constant \(K_{\mathrm{c}}\) at \(375^{\circ} \mathrm{C}\) is \(1.2 .\) Starting with \(\left[\mathrm{H}_{2}\right]_{0}=0.76 M,\left[\mathrm{~N}_{2}\right]_{0}=0.60 M,\) and \(\left[\mathrm{NH}_{3}\right]_{0}=\) \(0.48 M\), which gases will have increased in concentration and which will have decreased in concentration when the mixture comes to equilibrium?

Short Answer

Expert verified
The reaction shifts right to reach equilibrium, meaning the concentrations of Nitrogen gas and Hydrogen gas will decrease while the concentration of Ammonia will increase.

Step by step solution

01

Identification of the Reaction

The first step is to identify the reaction that is occurring. In this case, Nitrogen gas (N2) is reacting with Hydrogen gas (H2) to form Ammonia (NH3) and this reaction can be reversible. The given chemical equation perfectly represents this reaction process: \(N_{2}(g) + 3H_{2}(g) \leftrightarrow 2NH_{3}(g)\)
02

Calculation of Initial Reaction Quotient (Q)

The reaction quotient (Q) helps us to determine the direction of the reaction. Q can be calculated using the same formula as the equilibrium constant (Kc), but it uses the initial concentrations instead of equilibrium concentrations. We can calculate the reaction quotient (Q) using the initial concentrations provided in the exercise. Here, \(Q_c = \frac{[NH_{3}]^2_{initial}}{[N_{2}]_{initial}[H_{2}]_{initial}^3} = \frac{(0.48)^2}{(0.60)(0.76)^3} \)
03

Comparison

Compare the value of Qc with Kc. If Qc < Kc, the reaction will shift to the right to reach equilibrium, meaning the concentrations of N2 and H2 will decrease while the concentration of NH3 will increase because the forward reaction is favored. Conversely, if Qc > Kc, the reaction shifts to the left, meaning the concentration of NH3 decreases while the concentrations of N2 and H2 increase, since the reverse reaction is favored.
04

Interpreting the Outcome

Once you have calculated and compared the Qc value with the Kc, you can determine which gases will increase or decrease in concentration. You will find in this case Qc < Kc, indicating the reaction shifts to the right. Thus, the concentration of N2 and H2 will decrease and NH3 will increase as the reaction moves towards equilibrium.

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Key Concepts

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

Equilibrium Constant (Kc)
In chemical reactions that can occur in both forward and reverse directions, an equilibrium constant, denoted as Kc, provides vital information about the state of the reaction at equilibrium. It represents the ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their coefficients in the balanced equation.

The equilibrium constant is determined by the formula:
\[ K_c = \frac{[\text{Products}]^\text{coefficients}}{[\text{Reactants}]^\text{coefficients}} \]

For the ammonia synthesis reaction:\[ \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \],the Kc expression would be:\[ K_c = \frac{[\mathrm{NH}_{3}]^2}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^3} \].

At equilibrium, Kc is constant at a given temperature, indicating that the concentrations of reactants and products have reached a state of balance. If you alter the temperature, however, Kc will change, reflecting the temperature dependence of the equilibrium state.
Reaction Quotient (Qc)
What happens before a reaction reaches equilibrium? We look at the reaction quotient, Qc, to predict the direction in which a reaction mixture will proceed. Similar in form to Kc, the Qc uses the initial concentrations of reactants and products rather than the concentrations at equilibrium.

Using the given initial concentrations, the reaction quotient is calculated as follows:\[ Q_c = \frac{[\mathrm{NH}_{3}]^2_{initial}}{[\mathrm{N}_{2}]_{initial}[\mathrm{H}_{2}]_{initial}^3} \].

The value of Qc is then compared to Kc to determine which way the reaction will shift to reach equilibrium. If Qc is less than Kc (Qc < Kc), the forward reaction is not yet complete, and the concentration of products needs to increase while those of the reactants decrease. Conversely, if Qc is greater than Kc (Qc > Kc), the reverse reaction predominates, leading to a decrease in product concentration and an increase in reactants.
Le Chatelier's Principle
Le Chatelier's Principle offers a qualitative way to predict how a system at equilibrium responds to changes in concentration, temperature, or pressure. When a system at equilibrium faces a disturbance, it will shift to counteract the change and restore a new equilibrium.

For the ammonia synthesis example:\[ \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \],increasing the concentration of either reactants will shift the equilibrium toward the formation of more ammonia (NH3). Likewise, removing NH3 from the system will cause the reaction to shift right, increasing the production of NH3 to restore equilibrium.

Temperature changes will also affect the equilibrium. If the reaction in question is exothermic, increasing temperature will shift the equilibrium towards the reactants (to the left), while decreasing temperature favors the formation of products (shift right). This correlates to the concept that increased temperature adds energy into the system, which the endothermic reverse reaction could absorb, thus shifting the equilibrium backward.

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Most popular questions from this chapter

Consider the following reaction at equilibrium: $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ From the data shown here, calculate the equilibrium constant (both \(K_{P}\) and \(K_{\mathrm{c}}\) ) at each temperature. Is the reaction endothermic or exothermic? $$ \begin{array}{clr} \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & {[\mathrm{A}](M)} & {[\mathrm{B}](M)} \\ 200 & 0.0125 & 0.843 \\ 300 & 0.171 & 0.764 \\ 400 & 0.250 & 0.724 \end{array} $$

Eggshells are composed mostly of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) formed by the reaction $$\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)$$ The carbonate ions are supplied by carbon dioxide produced as a result of metabolism. Explain why eggshells are thinner in the summer when the rate of panting by chickens is greater. Suggest a remedy for this situation.

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00-L flask. At \(648 \mathrm{~K}\), there is \(0.0345 \mathrm{~mole}\) of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$

The following equilibrium constants have been determined for oxalic acid at \(25^{\circ} \mathrm{C}\) : $$\begin{array}{l}\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q) \\\\\qquad \begin{array}{r}K_{\mathrm{c}}^{\prime}=6.5 \times 10^{-2} \\\\\mathrm{HC}_{2} \mathrm{O}_{4}^{-}(a q) \Longrightarrow \mathrm{H}^{+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \\\K_{\mathrm{c}}^{\prime \prime}=6.1 \times 10^{-5}\end{array}\end{array}$$ Calculate the equilibrium constant for the following reaction at the same temperature: $$\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)$$

What effect does an increase in pressure have on each of the following systems at equilibrium? The temperature is kept constant and, in each case, the reactants are in a cylinder fitted with a movable piston. (a) \(\mathrm{A}(s) \rightleftharpoons 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftharpoons \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftharpoons \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\)

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