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Is a mixture of \(0.0205 \mathrm{mol} \mathrm{NO}_{2}(\mathrm{g})\) and \(0.750 \mathrm{mol}\) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) in a \(5.25 \mathrm{L}\) flask at \(25^{\circ} \mathrm{C},\) at equilibrium? If not, in which direction will the reaction proceed toward products or reactants? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3} \mathrm{at} 25^{\circ} \mathrm{C}$$

Short Answer

Expert verified
No, the system is not at equilibrium. The reaction will proceed in the forward direction, towards the products.

Step by step solution

01

Calculate the initial concentrations

Calculate the molar concentration of both \(NO_2\) and \(N_2O_4\) at the onset using the given quantity and volume in the flask. The concentration (in Molarity, M) is calculated by dividing the number of moles by the volume of the solution in liters. The initial concentration of \(NO_2\) is \(0.0205 mol / 5.25 L = 0.00390 M\) and for \(N_2O_4\) is \(0.750 mol / 5.25 L = 0.143 M\)
02

Calculate the reaction quotient \(Q_c\)

The reaction quotient \(Q_c\), similar to \(K_c\), gives the measure of the ratio of the concentrations of the products raised to their stoichiometric coefficient to that of the reactants raised to their stoichiometric coefficients, at any point in time. Here, \(Q_c\) is calculated via: \(Q_c = [NO_2]^2 / [N_2O_4]\). The efficent concentrations are used. Thus \(Q_c = (0.00390)^2 / 0.143 = 1.065x10^-4\)
03

Compare \(Q_c\) to \(K_c\)

Now, compare \(Q_c\) to \(K_c\). If \(Q_c < K_c\), this means the system is not at equilibrium and the reaction will proceed in the forward direction to reach equilibrium, converting more reactants to products. If \(Q_c > K_c\), this means the system is not at equilibrium and the reaction will proceed in the reverse direction to reach equilibrium, converting more products to reactants. Here, \(1.065x10^-4 < 4.61x10^-3\), so the reaction will proceed toward the products side.

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Key Concepts

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

Reaction Quotient
The Reaction Quotient, often denoted as \( Q_c \), is a vital concept in understanding chemical equilibrium. It measures a snapshot of a reaction at any given point in time, providing insight into whether a reaction is at equilibrium or not. To calculate \( Q_c \), we use the formula that involves concentrations of the products and reactants:
  • Numerator: Concentrations of the products, each raised to the power of its stoichiometric coefficient.
  • Denominator: Concentrations of the reactants, raised similarly.
In mathematical terms, for the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \), \( Q_c \) would be calculated as \( \frac{[NO_2]^2}{[N_2O_4]} \). By comparing this quotient to the Equilibrium Constant \( K_c \), we can determine the direction the reaction will proceed:
  • If \( Q_c < K_c \), the reaction moves forward, producing more products.
  • If \( Q_c > K_c \), the reaction shifts backward, forming more reactants.
Therefore, \( Q_c \) serves as the compass to balance the reaction's journey toward equilibrium.
Equilibrium Constant
The Equilibrium Constant, denoted as \( K_c \) when dealing with concentrations, is a crucial factor in determining the balance point of a chemical system at a given temperature. It illustrates the ratio of the concentration of products to reactants at equilibrium. For our reaction, \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \), we have \( K_c = 4.61 \times 10^{-3} \).
  • Each concentration in the expression is raised to the power of its respective stoichiometric coefficient.
  • Temperature-specific: Changing the temperature can alter \( K_c \).
  • An expression for \( K_c \) can be written as \( \frac{[NO_2]^2}{[N_2O_4]} \).
By comparing \( Q_c \) and \( K_c \):
  • If they are equal, the reaction is at equilibrium.
  • If not, they'll guide the system whether to favor the product or reactant side to reach equilibrium.
Thus, \( K_c \) not only helps verify an equilibrium state but also predicts how the system responds to changes.
Concentration
Concentration is a fundamental concept in chemistry, representing how much of a substance is present in a given volume. In the context of our chemical reaction, it’s crucial for determining both \( Q_c \) and \( K_c \).
  • Measured in moles per liter (Molarity, M).
  • Refers to the molar amount of a chemical species in a solution or gaseous state.
For our exercise:
  • Initial concentration of \( NO_2 \): calculated as \( \frac{0.0205}{5.25} = 0.00390 \) M\( \).
  • Initial concentration of \( N_2O_4 \): \( \frac{0.750}{5.25} = 0.143 \) M\( \).
These concentrations allow us to determine \( Q_c \) and eventually decide the direction in which the reaction moves to achieve equilibrium. Understanding concentration is crucial in calculating reaction quotients and predicting reaction dynamics. Remember, the precise adjustment of concentrations can shift the equilibrium and affect the extent to which reactants form products, continuing the journey towards equilibrium.

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Most popular questions from this chapter

Write equilibrium constant expressions, \(K_{\mathrm{p}},\) for the reactions (a) \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(1.06 \mathrm{g} \mathrm{H}_{2} \mathrm{S}\) in a 0.500 Lflask comes to equilibrium at \(1670 \mathrm{K}: 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{S}_{2}(\mathrm{g})\) found is \(8.00 \times 10^{-6}\) mol. Determine the value of \(K_{p}\) at 1670 K.

For the dissociation of \(\mathrm{I}_{2}(\mathrm{g})\) at about \(1200^{\circ} \mathrm{C}\) \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}), K_{\mathrm{c}}=1.1 \times 10^{-2} .\) What volume flask should we use if we want 0.37 mol I to be present for every \(1.00 \mathrm{mol} \mathrm{I}_{2}\) at equilibrium?

In one of Fritz Haber's experiments to establish the conditions required for the ammonia synthesis reaction, pure \(\mathrm{NH}_{3}(\mathrm{g})\) was passed over an iron catalyst at \(901^{\circ} \mathrm{C}\) and 30.0 atm. The gas leaving the reactor was bubbled through 20.00 mL of a HCl(aq) solution. In this way, the \(\mathrm{NH}_{3}(\mathrm{g})\) present was removed by reaction with HCl. The remaining gas occupied a volume of 1.82 L at STP. The \(20.00 \mathrm{mL}\) of \(\mathrm{HCl}(\mathrm{aq})\) through which the gas had been bubbled required \(15.42 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Another \(20.00 \mathrm{mL}\) sample of the same HCl(aq) through which no gas had been bubbled required \(18.72 \mathrm{mL}\) of \(0.0523 \mathrm{M} \mathrm{KOH}\) for its titration. Use these data to obtain a value of \(K_{\mathrm{p}}\) at \(901^{\circ} \mathrm{C}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})\)

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

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