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Chapter 1: Problem 19

$$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The reaction above came to equilibrium at a temperature of \(100^{\circ} \mathrm{C} .\) At equilibrium the partial pressure due to NOBr was 4 atmospheres, the partial pressure due to NO was 4 atmospheres, and the partial pressure due to \(\mathrm{Br}_{2}\) was 2 atmospheres. What is the equilibrium constant, \(K_{p},\) for this reaction at \(100^{\circ} \mathrm{C}\) ? (A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) 1 (D) 2

Short Answer

Expert verified
The equilibrium constant for the reaction, \( K_{p} \) is 2. So the answer is (D): 2

Step by step solution

01

Understanding the formula for equilibrium constant \( K_{p} \)

In general, for a reaction of the form \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K_{p} \) can be given as \( K_{p}=\frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}} \) where \( P_i \) stands for the partial pressure of the species \( i \) and \( a, b, c, d \) stand for their stoichiometric coefficients.
02

Identify the given values

From the question, we have \( P_{NOBr} = 4 \) atm, \( P_{NO} = 4 \) atm and \( P_{Br2} = 2 \) atm. We apply these values in the \( K_{p} \) formula.
03

Substitute the values in the formula

Substituting the values into the formula, we get: \( K_{p} = \frac{(P_{NO})^2(P_{Br2})}{(P_{NOBr})^2} = \frac{(4)^2(2)}{(4)^2} \)
04

Calculation of \( K_{p} \)

On simplifying the expression, we get \( K_{p} = \frac{32}{16} = 2 \)

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Most popular questions from this chapter

The value of \(K_{a}\) for \(\mathrm{HSO}_{4}^{-}\) is \(1 \times 10^{-2} .\) What is the value of \(K_{b}\) for \(\mathrm{SO}_{4}^{2-} ?\) (A) \(K_{b}=1 \times 10^{-12}\) (B) \(K_{b}=1 \times 10^{-8}\) (C) \(K_{b}=1 \times 10^{-2}\) (D) \(K_{b}=1 \times 10^{2}\)

A laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer? (A) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad K_{a}=5.9 \times 10^{-2}\) (B) \(\mathrm{H}_{3} \mathrm{AsO}_{4} \quad K_{a}=5.6 \times 10^{-3}\) (C) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \quad K_{a}=1.8 \times 10^{-5}\) (D) \(\mathrm{HOCl}\) \(\quad K_{a}=3.0 \times 10^{-8}\)

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . What are the concentrations of \(\mathrm{Sr}^{2+}\) and \(\mathrm{F}^{-}\) in the beaker? (A) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (B) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\) (C) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (D) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} \mathrm{M}\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\)

Which of the following species is amphoteric? (A) \(\mathrm{H}^{+}\) (B) \(\mathrm{CO}_{3}^{2-}\) (C) \(\mathrm{HCO}_{3}^{-}\) (D) \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

\(\begin{array}{ll}{\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}} & {E^{\circ}=+0.3 \mathrm{V}} \\ {\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}} & {E^{\circ}=-0.8 \mathrm{V}} \\ {\mathrm{Mn}^{2+}+2 e^{-} \rightarrow \mathrm{Mn}} & {E^{\circ}=-1.2 \mathrm{V}}\end{array}\) Based on the reduction potentials given above, which of the following reactions will be favored? (A) \(\mathrm{Mn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Mn}+\mathrm{Cu}^{2+}\) (B) \(\mathrm{Mn}^{2+}+\mathrm{Zn} \rightarrow \mathrm{Mn}+\mathrm{Zn}^{2+}\) (C) \(\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}\) (D) \(\mathrm{Zn}^{2+}+\mathrm{Mn} \rightarrow \mathrm{Zn}+\mathrm{Mn}^{2+}\)
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