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Nitric oxide (NO) reacts readily with chlorine gas as follows: $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)$$ At \(700 \mathrm{K},\) the equilibrium constant \(K_{p}\) for this reaction is \(0.26 .\) Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium. (a) \(P_{\mathrm{NO}}=0.15\) atm \(, P_{\mathrm{Cl}_{2}}=0.31 \mathrm{atm}, P_{\mathrm{NOCl}}=0.11 \mathrm{atm}\) (b) \(R_{\mathrm{NO}}=0.12 \mathrm{atm}, P_{\mathrm{Cl}_{2}}=0.10 \mathrm{atm}, P_{\mathrm{NOCl}}=0.050 \mathrm{atm}\) (c) \(R_{\mathrm{NO}}=0.15 \mathrm{atm}, P_{\mathrm{Cl}_{2}}=0.20 \mathrm{atm,}\) \(R_{\mathrm{NOC}}=5.10 \times 10^{-3} \mathrm{atm}\)

Short Answer

Expert verified
(a) The reaction needs to produce more products (NOCl) to reach equilibrium. (b) The reaction needs to produce more reactants (NO and Cl₂) to reach equilibrium. (c) The reaction needs to produce more products (NOCl) to reach equilibrium.

Step by step solution

01

Write the expression for the reaction quotient (Qp)

We can write the expression for Qp using the partial pressures of the reactants and products: \[Q_p = \frac{P_{NOCl}^2}{P_{NO}^2 \cdot P_{Cl_2}}\] The subscript of each species represents its partial pressure.
02

Calculate Qp for part (a)

We can calculate the value of Qp by substituting the given values into the formula: \[Q_p = \frac{(0.11\;atm)^2}{(0.15\;atm)^2 \cdot (0.31\;atm)}\] \[Q_p = 0.1709\]
03

Compare Qp and Kp for part (a)

Given that \(K_p = 0.26\), since \(Q_p < K_p\), the reaction should proceed in the forward direction to reach equilibrium. The mixture will need to produce more products (NOCl) to reach equilibrium.
04

Calculate Qp for part (b)

After substituting the given values into the formula, we get: \[Q_p = \frac{(0.050\;atm)^2}{(0.12\;atm)^2 \cdot (0.10\;atm)}\] \[Q_p = 0.3472\]
05

Compare Qp and Kp for part (b)

Since \(Q_p > K_p\) (\(0.3472 > 0.26\)), the reaction will need to proceed in the reverse direction, producing more reactants (NO and Cl₂) to reach equilibrium.
06

Calculate Qp for part (c)

Putting in the given values into the formula, we find: \[Q_p = \frac{(5.10 \times 10^{-3}\;atm)^2}{(0.15\;atm)^2 \cdot (0.20\;atm)}\] \[Q_p = 0.000567\]
07

Compare Qp and Kp for part (c)

Here, since \(Q_p < K_p\) (\(0.000567 < 0.26\)), the reaction should proceed in the forward direction to reach equilibrium, producing more products (NOCl). In conclusion, For part (a), the reaction needs to produce more products to reach equilibrium. For part (b), the reaction needs to produce more reactants to reach equilibrium. For part (c), the reaction needs to produce more products to reach equilibrium.

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Key Concepts

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

Understanding the Reaction Quotient Qp
In chemical processes, it's essential to determine the direction in which a reaction will proceed to achieve equilibrium. The reaction quotient, denoted as Qp, serves this purpose in the context of gaseous reactions where pressure is a key factor. By comparing the partial pressures of the reactants and products at any point during the reaction, Qp helps predict the system's behavior.

For the reaction of nitric oxide (NO) with chlorine gas (Cl₂) to form nitrosyl chloride (NOCl), we use the expression \(Q_p = \frac{P_{NOCl}^2}{P_{NO}^2 \times P_{Cl_2}}\) to calculate Qp. This quotient reflects the ratio of products to reactants at a given moment. If Qp is less than the equilibrium constant Kp, the reaction will shift towards products, as seen in parts (a) and (c) of the example. Conversely, if Qp exceeds Kp, the reaction will favor the formation of reactants to reach equilibrium, as demonstrated in part (b).

Subtle changes in the pressures of the involved gases significantly influence the system's behavior. By computing and comparing Qp with the known Kp value, we can predict whether the reaction mixture is at equilibrium or if changes will occur to reach a state of balance.
Equilibrium Constant Kp and Its Role
The equilibrium constant, Kp, is a central concept in understanding chemical equilibria. It's a unique value for a given chemical reaction at a constant temperature, reflecting the ratio of the concentrations (or partial pressures in the case of gases) of the products to reactants when the reaction is at equilibrium. For the reaction \(2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)\), the given Kp value of 0.26 at 700 K provides a reference point to determine the state of the reaction.

When a reaction system's Qp value is compared to Kp, it reveals whether the reaction is already at equilibrium or which direction it needs to proceed to attain equilibrium. A Qp lower than Kp, indicates an excess of reactants, and thus the reaction will shift to form more products. In contrast, a Qp higher than Kp signifies an excess of products, pushing the reaction towards forming more reactants. In essence, Kp is a guiding star for the reaction's journey towards equilibrium, giving a clear-cut indication of the reaction’s progress based on the current conditions.
Le Chatelier's Principle in Action
Le Chatelier's Principle is a powerful tool for predicting how a reaction system at equilibrium responds to changes in conditions such as concentration, pressure, and temperature. Essentially, it states that if an external change is applied to a system at equilibrium, the system adjusts in a way that partially counteracts the change.

Applying this principle to the reaction between NO and Cl₂, we can infer that increasing the pressure of either reactant or decreasing the pressure of the product would drive the reaction to produce more NOCl. Conversely, reducing the pressure of the reactants or increasing the pressure of NOCl will shift the equilibrium in favor of the reactants to restore balance.

Moreover, temperature changes play a critical role; for an exothermic reaction, an increase in temperature shifts the equilibrium towards the reactants, behaving as if heat is a product. Conversely, for endothermic reactions, higher temperatures favor the formation of products, as if heat behaves like a reactant. These relationships provide a framework for understanding the complex dynamics of equilibrium and how a system adapts to maintain balance, a concept instrumental in fields ranging from industrial chemistry to environmental science.

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Most popular questions from this chapter

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{4}\) at 700 \(\mathrm{K} .\) In a \(2.00-\mathrm{L}\) vessel, the equilibrium mixture contains 1.17 \(\mathrm{g}\) of \(\mathrm{SO}_{3}\) and 0.105 \(\mathrm{g}\) of \(\mathrm{O}_{2} .\) How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each \(\mathrm{Hb}\) can bind 4 \(\mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which 50\(\%\) of the protein is saturated. Fetal hemoglobin has a P50 value of 19 torr, and adult hemoglobin has a \(\mathrm{P} 50\) value of 26.8 torr. Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) in a fetus, compared to \(K_{c}\) for the same reaction in an adult.

A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug molecules bind the protein in a \(1 : 1\) ratio to form a drug-protein complex. The protein concentration in aqueous solution at \(25^{\circ} \mathrm{C}\) is \(1.50 \times 10^{-6} \mathrm{M}\) . Drug \(\mathrm{A}\) is introduced into the protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\) Drug \(\mathrm{B}\) is introduced into a separate, identical protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\) . At equilibrium, the drug \(\mathrm{A}-\) protein solution has an \(\mathrm{A}\) -protein complex concentration of \(1.00 \times 10^{-6} \mathrm{M},\) and the drug \(\mathrm{B}\) solution has a B-protein complex concentration of \(1.40 \times 10^{-6} \mathrm{M}\) Calculate the \(K_{c}\) value for the A-protein binding reaction and for the B-protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

\(\mathrm{At} 900 \mathrm{K},\) the following reaction has \(K_{p}=0.345 :\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(S \mathrm{O}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

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