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At \(460^{\circ} \mathrm{C},\) the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g) $$ has \(K_{\mathrm{c}}=85.0\). A reaction flask at \(460^{\circ} \mathrm{C}\) contains these gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]=0.00250 \mathrm{M}\), \(\left[\mathrm{NO}_{2}\right]=0.00350 \quad M,[\mathrm{NO}]=0.0250 \quad M,\) and \(\left[\mathrm{SO}_{3}\right]=0.0400 \mathrm{M}\) (a) Is the reaction at equilibrium? (b) If not, in which direction will the reaction proceed to arrive at equilibrium?

Short Answer

Expert verified
The reaction is not at equilibrium because \(Q > K_c\). The reaction will proceed in the reverse direction to reach equilibrium.

Step by step solution

01

- Write Down Equilibrium Constant Expression

For the given reaction, the equilibrium constant expression is given by: \[ K_c = \frac{[\mathrm{NO}][\mathrm{SO}_3]}{[\mathrm{SO}_2][\mathrm{NO}_2]} \]
02

- Calculate the Reaction Quotient (Q)

Using the given concentrations, calculate the reaction quotient, Q, to compare with the equilibrium constant, Kc. \[ Q = \frac{[\mathrm{NO}][\mathrm{SO}_3]}{[\mathrm{SO}_2][\mathrm{NO}_2]} = \frac{(0.0250\ M)(0.0400\ M)}{(0.00250\ M)(0.00350\ M)} \]
03

- Evaluate the Reaction Quotient (Q)

Evaluate Q to determine if the reaction is at equilibrium. \[ Q = \frac{(0.0250)(0.0400)}{(0.00250)(0.00350)} = \frac{0.0010}{0.00000875} = 114.29 \]
04

- Compare Q and Kc

Compare the calculated Q to the given Kc to determine the direction of the reaction. Since \(Q > K_c\), the reaction will proceed in the reverse direction to reach equilibrium.

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Key Concepts

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

Equilibrium Constant (Kc)
The equilibrium constant, denoted as Kc, is a fundamental concept in chemical equilibrium. It quantifies the ratio of concentration of products to reactants at equilibrium, raised to their stoichiometric coefficients from the balanced equation. For the given reaction \[\mathrm{SO}_{2}(g) + \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{NO}(g) + \mathrm{SO}_{3}(g)\] the equilibrium constant expression would be written as \[K_c = \frac{[\mathrm{NO}][\mathrm{SO}_3]}{[\mathrm{SO}_2][\mathrm{NO}_2]}\].

The value of Kc provides critical insight into the reaction's characteristics at a given temperature. A high Kc value indicates a greater concentration of products at equilibrium, suggesting the reaction goes nearly to completion. Conversely, a low Kc implies a higher concentration of reactants at equilibrium, meaning the reaction scarcely proceeds.
Reaction Quotient (Q)
The reaction quotient, Q, is a snapshot of a reaction's status at any particular moment, giving the same ratio of product and reactant concentrations as the Kc expression, but not necessarily at equilibrium. It is calculated using the formula \[Q = \frac{[\mathrm{NO}][\mathrm{SO}_3]}{[\mathrm{SO}_2][\mathrm{NO}_2]}\] using current concentrations. By comparing Q to Kc, we determine the system's direction towards equilibrium:
  • If Q = Kc, the system is at equilibrium.
  • If Q > Kc, the reaction will shift towards reactants to reach equilibrium.
  • If Q < Kc, the reaction will shift towards products to reach equilibrium.
In our example, Q was calculated to be larger than Kc, indicating an excess of products and therefore the reaction will shift towards reactants to achieve equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle provides a qualitative approach to predicting the effect of changes in concentration, temperature, or pressure on a system at equilibrium. According to this principle, if any of these conditions are altered, the equilibrium will shift to counteract the imposed change.

For instance, if the concentration of a product is increased, the equilibrium will shift towards reactants to reduce this change. Similarly, if the temperature of an exothermic reaction increases, the reaction will shift to produce more reactants, absorbing the excess heat. Le Chatelier's Principle helps us to understand why the reaction quotient Q being different than Kc in our scenario prompts the reaction to proceed in the reverse direction to re-establish equilibrium.
Equilibrium Concentration Calculations
Calculating equilibrium concentrations involves using the equilibrium constant expression in conjunction with the amounts of reactants and products present. Start by writing the balanced equation for the chemical reaction and the expression for Kc. Then, determine the concentrations of each species involved. In some cases, it may be necessary to perform an 'ICE' table analysis (Initial, Change, Equilibrium) to understand how the concentrations shift from initial conditions to equilibrium.

In the given problem, the initial concentrations were used to calculate Q, and since it was different from Kc, we know that the reaction is not at equilibrium. To find the new equilibrium concentrations, adjustments need to be made in the direction predicted by comparing Q and Kc. This involves a combination of stoichiometry, algebra, and the equilibrium constant expression, ensuring that the proportions of reactants and products adhere to the balanced chemical equation.

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Most popular questions from this chapter

At a certain temperature, \(K_{\mathrm{c}}=0.18\) for the equilibrium $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ If \(0.026 \mathrm{~mol}\) of \(\mathrm{PCl}_{5}\) is placed in a \(2.00 \mathrm{~L}\) vessel at this temperature, what will the concentration of \(\mathrm{PCl}_{3}\) be at equilibrium?

Why do we omit the concentrations of pure liquids and pure solids from the mass action expression of heterogeneous reactions?

Write the equilibrium law for each of the following gas phase reactions in terms of molar concentrations. (a) \(3 \mathrm{Cl}_{2}(g)+\mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{NCl}_{3}(g)+3 \mathrm{HCl}(g)\) (b) \(\mathrm{PCl}_{3}(g)+\mathrm{PBr}_{3}(g) \rightleftharpoons \mathrm{PCl}_{2} \operatorname{Br}(g)+\mathrm{PClBr}_{2}(g)\) (c) \(\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HNO}_{2}(g)\) (d) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)\) (e) \(\mathrm{Br}_{2}(g)+5 \mathrm{~F}_{2}(g) \rightleftharpoons 2 \mathrm{BrF}_{5}(g)\)

The heterogeneous reaction \(2 \mathrm{HCl}(g)+\mathrm{I}_{2}(s) \rightleftharpoons\) \(2 \mathrm{HI}(g)+\mathrm{Cl}_{2}(g)\) has \(K_{\mathrm{c}}=1.6 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). Suppose \(0.100 \mathrm{~mol}\) of \(\mathrm{HCl}\) and solid \(\mathrm{I}_{2}\) are placed in a \(1.00 \mathrm{~L}\) container. What will be the equilibrium concentrations of HI and \(\mathrm{Cl}_{2}\) in the container?

How will the value of \(K_{\mathrm{P}}\) for the following reactions be affected by an increase in temperature? (a) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) $$ \Delta H^{\circ}=-18 \mathrm{~kJ} $$ (b) \(\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons 3 \mathrm{NO}(g)\) $$ \Delta H^{\circ}=+155.7 \mathrm{~kJ} $$ (c) \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) \(\Delta H^{\circ}=-77.07 \mathrm{~kJ}\)

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