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Chapter 15: Problem 51

At \(2000^{\circ} \mathrm{C},\) the equilibrium constant for the reaction $$2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ is \(K_{c}=2.4 \times 10^{3} .\) If the initial concentration of \(\mathrm{NO}\) is \(0.175 \mathrm{M},\) what are the equilibrium concentrations of \(\mathrm{NO}\) \(\mathrm{N}_{2},\) and \(\mathrm{O}_{2} ?\)

Short Answer

Expert verified
The equilibrium concentrations are: [NO] = 0.105 M, [N2] = 0.035 M, and [O2] = 0.035 M.

Step by step solution

01

Write the equilibrium constant expression

For the given reaction: \( 2 NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g) \) The equilibrium constant expression is given by: \( K_c = \frac{[N_{2}][O_{2}]}{[NO]^2} \)
02

Create an ICE table (Initial, Change, Equilibrium)

Organize the information in an ICE table: \[ \begin{array}{c|c|c|c} \text{Species} & \text{Initial [M]} & \text{Change [M]} & \text{Equilibrium [M]} \\ \hline NO(g) & 0.175 & -2x & 0.175-2x \\ N_{2}(g) & 0 & x & x \\ O_{2}(g) & 0 & x & x \end{array} \] In the table above, x represents the change in moles when the reaction reaches equilibrium.
03

Substitute the ICE table values into the equilibrium constant expression

Write the equilibrium constant expression using the equilibrium concentration values in the ICE table: \( K_c = \frac{x^2}{(0.175 - 2x)^2} \) Given, \( K_c = 2.4 \times 10^{3} \). Now substitute the value of Kc into the expression: \( 2.4 \times 10^{3} = \frac{x^2}{(0.175 - 2x)^2} \)
04

Solve for x

To solve for x, we will first rearrange the equation and then use a suitable method for solving the equation: \( x^2 = (2.4 \times 10^{3}) (0.175 - 2x)^2 \) This is a quadratic equation in x. You can simplify the equation and apply a quadratic formula, or solve by using any method you prefer (like graphing or iteration). Upon solving, we get: \(x = 0.035\)
05

Find the equilibrium concentrations

Using the value of x, find the equilibrium concentrations of each species from the ICE table: \( [NO]_{eq} = 0.175 - 2(0.035) = 0.105\,M \) \( [N_{2}]_{eq} = 0 + 0.035 = 0.035\,M \) \( [O_{2}]_{eq} = 0 + 0.035 = 0.035\,M \) Thus, the equilibrium concentrations are: [NO] = 0.105 M, [N2] = 0.035 M, and [O2] = 0.035 M.

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Most popular questions from this chapter

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{S}(g) \quad K_{c}=1.08 \times 10^{7} \mathrm{at} 700^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\) . (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{S} ?(\mathbf{c})\) Calculate the value of \(K_{c}\) if you rewrote the equation \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{S}(g)\)

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?

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? $$\begin{array}{ll}{\text { (a) } 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)} & {K_{p}=5.0 \times 10^{12}} \\\ {\text { (b) } 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)} & {K_{c}=5.8 \times 10^{-18}}\end{array}$$

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? $$\begin{array}{ll}{\text { (a) } \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)} & {K_{c}=1.5 \times 10^{-10}} \\ {\text { (b) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)} & {K_{p}=2.5 \times 10^{9}}\end{array}$$
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