The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at 1000 \(\mathrm{K}\) . (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2},\) or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) (c) Calculate \(K_{c}\) for \(\operatorname{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\)

Short Answer

Expert verified
(a) At 1000 K, the equilibrium favors the reactants NO and Br2 since Kc < 1. (b) The Kc for the reverse reaction is 76.9. (c) The Kc for the new reaction with halved product coefficient is 0.114.

Step by step solution

01

(a) Determine which side is favored in the equilibrium

First, let's look at the given reaction and its equilibrium constant: \[2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\] \[K_{c} = 1.3 \times 10^{-2}\] The equilibrium constant, Kc, represents the ratio of the concentrations of the products to the reactants. If Kc >> 1, products are favored, and if Kc << 1, reactants are favored. Here, Kc is smaller than 1, so the equilibrium will favor the reactants, which are NO and Br2.
02

(b) Calculate Kc for the reverse reaction

Given that Kc for the original reaction is 1.3 x 10^-2, we will find the Kc for the reverse reaction: \[2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\] The relationship between the equilibrium constants of forward and reverse reactions is given by: \[K_c^{'(reverse)} = \frac{1}{K_c^{(forward)}}\] Substitute the given value: \[K_c^{(reverse)} = \frac{1}{1.3 \times 10^{-2}}\] Now, calculate Kc for the reverse reaction: \[K_c^{(reverse)} = 76.9\]
03

(c) Calculate Kc for the new reaction with halved product coefficient

The new reaction is: \[\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\] First, let's rewrite the original reaction by dividing each coefficient by 2: \[\mathrm{NO}(g) + \frac{1}{2}\mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{NOBr}(g)\] Now, to find the Kc for this new reaction, we can use the formula: \[K_c^{(new)} = \sqrt[K_c^{(original)}]{n}\] where n is the new coefficient and Kc (original) is the given Kc for the original reaction. Here, n = 2 (this is because the new reaction is created by halving the coefficients of all species in the initial reaction). \[K_c^{(new)} = \sqrt[1.3 \times 10^{-2}]{2}\] Now, calculate Kc for the new reaction: \[K_c^{(new)} = 0.114\] So, the equilibrium constants for the reactions are: (a) Kc (original) = 1.3 x 10^-2 (favoring NO and Br2) (b) Kc (reverse) = 76.9 (c) Kc (new) = 0.114

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry that occurs when the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in no overall change in the concentration of reactants and products over time. In a chemical equation, we depict this by the double arrows pointing in opposite directions, as shown in the reaction
\(2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g)\).
at a certain temperature. At equilibrium, the concentrations remain constant but are not necessarily equal.

An equilibrium constant (\(K_c\)) value gives us an insight into the composition of the equilibrium mixture. A low \(K_c\) value (much less than 1) signifies that, upon reaching equilibrium, the reactants are favored, meaning their concentrations will remain comparatively high, while high \(K_c\) values (much greater than 1) indicate a product-favored equilibrium, where the products have a higher concentration. In our exercise, a \(K_c = 1.3 \times 10^{-2}\) tells us that at 1000 K, the reactants, NO and \(\text{Br}_2\), are favored over the product NOBr.
Reaction Quotient
The reaction quotient (\(Q\)) is a measure that tells us the direction in which a reaction is likely to proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant \(K_c\), but with the initial concentrations of reactants and products rather than their equilibrium concentrations.

When analyzing the reaction quotient, we consider:
  • If \(Q = K_c\), the system is at equilibrium.
  • If \(Q < K_c\), the reaction proceeds forward to produce more products until equilibrium is reached.
  • If \(Q > K_c\), the reaction proceeds in reverse to produce more reactants until equilibrium is established.
In the textbook exercise provided, the reaction quotient is not explicitly calculated, but understanding its role helps us appreciate how the system shifts to reach equilibrium. If we had initial concentrations, we could compare \(Q\) to the provided \(K_c\) to predict the direction of the reaction.
Le Chatelier's Principle
Le Chatelier's Principle is a qualitative tool that predicts how a change in conditions (such as concentration, pressure, or temperature) can affect the position of equilibrium. Essentially, if a system at equilibrium is disturbed by a change in conditions, the system responds by adjusting in a way that counteracts the change and re-establishes equilibrium.

This principle can be understood through different changes:
  • Increasing the concentration of reactants will shift the equilibrium to the right, favoring product formation,
  • Increasing the concentration of products shifts it to the left, favoring reactants,
  • Changes in pressure for gaseous reactions also shift the equilibrium depending on the mole change in the reaction, and
  • Temperature effects depend on whether the reaction is exothermic or endothermic.
For instance, if the concentration of \(NO\) or \(\text{Br}_2\) is increased for the given reaction, Le Chatelier's Principle predicts a shift toward the right to form more NOBr, counteracting the concentration change. Such a shift is indicated without altering the constants (\(K_c\) and \(Q\)), which only change with temperature.

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Most popular questions from this chapter

NiO is to be reduced to nickel metal in an industrial process by use of the reaction $$\mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g)$$ At \(1600 \mathrm{K},\) the equilibrium constant for the reaction is \(K_{p}=6.0 \times 10^{2} .\) If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur?

If \(K_{c}=1\) for the equilibrium \(2 \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g),\) what is the relationship between \([\mathrm{A}]\) and \([\mathrm{B}]\) at equilibrium?

Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C}\) , where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .\) At equilibrium it is found that \([\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{I}_{2}\right]=4.79 \times 10^{-4} \mathrm{M} .\) What is the value of \(K_{c}\) at this temperature?

The following equilibria were measured at 823 K: \begin{equation} \begin{aligned} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) & \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{aligned} \end{equation} (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction \(\operatorname{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at 823 \(\mathrm{K}\) . (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{K} ?\) (c) If you were to place 5.00 \(\mathrm{g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of 250 \(\mathrm{mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of 1.00 atm and a temperature of \(298 \mathrm{K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to 823 \(\mathrm{K}\) and allowed to come to equilibrium, how much \(\mathrm{CoO}(s)\) remains?

At \(25^{\circ} \mathrm{C},\) the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4} .\) What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) in a saturated solution of \(\mathrm{CaCrO}_{4} ?\)

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