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The equilibrium constant \(K_{c}\) for the reaction \(\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)\) is \(16 .\) If 1 mole of each of all the four gases is taken in \(1 \mathrm{dm}^{3}\) vessel, the equilibrium concentration of NO would be : (a) \(0.4 \mathrm{M}\) (b) \(0.6 \mathrm{M}\) (c) \(1.4 \mathrm{M}\) (d) \(1.6 \mathrm{M}\)

Short Answer

Expert verified
The equilibrium concentration of NO would be 1.6 M.

Step by step solution

01

Write the expression for the equilibrium constant

The equilibrium constant expression for the given reaction \(\mathrm{SO}_{2}(g) + \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) + \mathrm{NO}(g)\) is \(K_{c} = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]}\).
02

Set up the initial concentrations and change in concentration

Consider the initial concentration of each gas to be 1 M, since we have 1 mole of each gas in a 1 dm^3 vessel. Then set up a table with the initial concentrations, the change in concentration, and the equilibrium concentrations using variables.
03

Calculate the equilibrium concentrations

If x is the change in concentration of \(SO_{2}\) and \(NO_{2}\) at equilibrium, then the concentration of \(SO_{3}\) and \(NO\) will be increased by x. Therefore, the concentration of \(SO_{3}\) will be \(1+ x\) M and that of \(NO\) will be \(1 + x\) M at equilibrium.
04

Substitute the equilibrium concentrations into the equilibrium constant expression

Substitute the equilibrium concentrations back into the equilibrium expression and solve for x. The equilibrium constant is given as 16, so\(16 = \frac{(1+x)(1+x)}{(1-x)(1-x)}\).
05

Solve for x

Solve the quadratic equation for x, keeping in mind that x cannot be greater than 1 since the concentration of gases cannot be negative.
06

Calculate the equilibrium concentration of NO

Once x is found, use it to calculate the equilibrium concentration of \(\mathrm{NO}\) which is \(1 + x\).

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Key Concepts

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

Chemical Equilibrium
In chemistry, chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction, meaning there's no net change in the amounts of reactants and products over time. This dynamic balance does not imply equal concentrations of reactants and products, but rather that their concentrations have stabilized to a constant ratio.

Understanding equilibrium is fundamental for predicting how a system behaves when conditions change, such as concentration, temperature, and pressure. The equilibrium state can be altered by changing these conditions, resulting in a shift of the equilibrium position according to Le Châtelier's principle. This principle posits that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
Equilibrium Concentration Calculation
When dealing with a chemical reaction at equilibrium, equilibrium concentration calculations are essential for quantifying the concentration of reactants and products. It involves setting up and solving a system of equations based on the balanced chemical equation and the initial concentrations of the substances involved. These calculations often use ice tables (Initial, Change, Equilibrium) to organize and simplify the calculation process.

For instance, suppose we have a reaction with initial concentrations and need to find the concentration of a reactant or product at equilibrium. The change in concentration due to the reaction is denoted by a variable, say 'x'. The 'ice' method allows us to express equilibrium concentrations in terms of x, subsequently using the equilibrium constant to solve for this variable, and ultimately finding the concentration of interest.
Equilibrium Constant Expression
The equilibrium constant expression is a mathematical representation of a chemical equilibrium. It is denoted as Kc when concentrations are in moles per liter (molarity) and Kp when using partial pressures. The general form of the expression for a reaction is given by:

Kc = \(\frac{[products]}{[reactants]}\),where the concentrations of the products and reactants are raised to the power of their stoichiometric coefficients. Products form the numerator and reactants, the denominator. It's important to note that only gases and aqueous solutions are included in the expression, excluding pure solids and liquids.

In practice, the equilibrium constant expression helps predict the direction in which a reaction will proceed to reach equilibrium, and by calculating its value, chemists can understand the extent to which a reaction will favor products or reactants under certain conditions.

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Most popular questions from this chapter

For the reversible reaction, \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) at \(500^{\circ} \mathrm{C}\), the value of \(K_{p}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmospheres. The corresponding value of \(K_{c}\) with concentration in mole litre \(^{-1}\), is: (a) \(1.44 \times 10^{-5} /(0.082 \times 500)^{-2}\) (b) \(1.44 \times 10^{-5} /(8.314 \times 773)^{-2}\) (c) \(1.44 \times 10^{-5} /(0.082 \times 773)^{2}\) (d) \(1.44 \times 10^{-5} /(0.082 \times 773)^{-2}\)

A pressure cooker reduces cooking time for food because: (a) the higher pressure inside the cooker crushes the food material (b) cooking involves chemical changes helped by a rise in temperature (c) heat is more evenly distributed in the cooking space (d) boiling point of water involved in cooking is increased

The most stable oxides of nitrogen will be : (a) \(2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) ; \quad K=6.7 \times 10^{16} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+5 \mathrm{O}_{2}(g) ; \quad K=1.2 \times 10^{24} \mathrm{~mol}^{5} \mathrm{~L}^{-5}\) (c) \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)\) \(K=2.2 \times 10^{30}\) (d) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) ; \quad K=3.5 \times 10^{33} \mathrm{~mol} \mathrm{~L}^{-1}\)

Consider the following gaseous equilibria given below : (I) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\); Eqm. Constant \(=K_{1}\) (II) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} ; \quad\) Eqm. Constant \(=K_{2}\) (III) \(\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O} ;\) Eqm. Constant \(=K_{3}\) The equilibrium constant for the reaction, \(2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}\) in terms of \(K_{1}, K_{2}\) and \(K_{3}\) will be : \(\begin{array}{llll}\text { (a) } K_{1} K_{2} K_{3} & \text { (b) } \frac{K_{1} K_{2}}{K_{3}} & \text { (c) } \frac{K_{1} K_{3}^{2}}{K_{2}} & \text { (d) } \frac{K_{2} K_{3}^{3}}{K_{1}}\end{array}\)

For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \quad \Delta H=-93.6 \mathrm{~kJ} \mathrm{~mol}^{-1} $$ the number of moles of \(\mathrm{H}_{2}\) at equilibrium will increase if : (a) volume is increased (b) volume is decreased (c) argon gas is added at constant volume (d) \(\mathrm{NH}_{3}\) is removed

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