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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 NO is \(0.250 \mathrm{M}\), what are the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) ?

Short Answer

Expert verified
At equilibrium, the concentrations for the given reaction are approximately \(\mathrm{[NO]} \approx 0.098\:M\), \(\mathrm{[N}_{2}] \approx 0.076\:M\), and \(\mathrm{[O}_{2}] \approx 0.076\:M\).

Step by step solution

01

Write down the initial concentrations

Initially, we are given the concentration of \(\mathrm{NO}\) which is 0.250 M. The concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are not given, so we can assume they are initially 0 M. \[\mathrm{NO} \: \: (0.250\: M) \] \[\mathrm{N}_{2} \: \: (0 \: M)\] \[\mathrm{O}_{2} \: \: (0 \: M)\]
02

Set up the ICE table

Using the ICE table, we will write down the changes in the concentrations as the reaction proceeds: \( %% \begin{array}{c|c} & \mathrm{[NO]} & \mathrm{[N}_{2}] & \mathrm{[O}_{2}]\\ \hline \text{Initial} & 0.250 & 0 & 0\\ \text{Change} & -2x & x & x\\ \text{Equilibrium} & 0.250-2x & x& x \end{array} %%"\ \)
03

Write the equilibrium constant expression

We are given \(K_c = 2.4 \times 10^3\). Then, let's write down the equilibrium constant expression using the equilibrium concentrations from the ICE table: \[K_c = \frac{\mathrm{[N}_{2}\mathrm{]}\mathrm{[O}_{2}\mathrm{]}}{\mathrm{[NO]^2}}\]
04

Substitute the equilibrium concentrations and solve for x

Now we substitute the equilibrium concentrations from the ICE table and solve for x: \[ 2.4 \times 10^3 = \frac{x \cdot x}{(0.250 - 2x)^2} \] Solve for x (let the solver find the positive root of the equation, since concentrations should be positive): \(x \approx 0.076\)
05

Find the equilibrium concentrations

Now that we have the value of x, we can plug it back into the equilibrium concentrations from the ICE table: \[\mathrm{[NO]_{eq}} = 0.250 - 2x \approx \] \[\mathrm{[N}_{2}]_{eq} = x \approx \] \[\mathrm{[O}_{2}]_{eq} = x \approx \] The equilibrium concentrations are as follows: \[\mathrm{[NO]} \approx 0.098 \: M\] \[\mathrm{[N}_{2}] \approx 0.076 \: M\] \[\mathrm{[O}_{2}] \approx 0.076 \: M\]

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Most popular questions from this chapter

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (a) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

The following equilibria were attained at \(298 \mathrm{~K}:\) $$\begin{array}{c} \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s) \quad K_{c}=5.6 \times 10^{9} \\ \mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(a q) \\ K_{c}=1.6 \times 10^{7}\end{array}$$ Based on these equilibria, calculate the equilibrium constant $\text { for } \mathrm{AgCl}(s)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}(a q)+\mathrm{Cl}^{-}(a q)$ at \(298 \mathrm{~K}\).

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

At \(700 \mathrm{~K}\), the equilibrium constant for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=77\). A flask is charged with \(202.7 \mathrm{kPa}\) of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at \(700 \mathrm{~K}\). (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?\) (b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Calculate \(K_{c}\) at \(900 \mathrm{~K}\) for $2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\( if \)K_{p}=0.0572$ at this temperature.
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