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Chapter 13: Problem 101

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ If \(2.0\) moles of \(\mathrm{NO}\) and \(1.0\) mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Short Answer

Expert verified
The equilibrium concentrations of all species in the given reaction are approximately: - \([\mathrm{NOCl}] \approx 7.98\times 10^{-3}M\) - \([\mathrm{NO}] \approx 1.984M\) - \([\mathrm{Cl}_{2}] \approx 0.992M\)

Step by step solution

01

Write down the balanced chemical equation

We are given the balanced chemical equation: \[ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Cl}_{2}(g)\]
02

Set up the ICE table

Set up the ICE table using the given initial concentrations and changes in concentration: | | NOCl |⟷ | 2NO | + | Cl2 | |-----------|------|--|-----|--|-----| | Initial | 0 | | 2.0 | | 1.0 | | Change | +x | | -2x | | -x | | Equilibrium| x | | 2.0 - 2x | | 1.0 - x | Where: - x is the concentration of NOCl at equilibrium - 2.0 - 2x is the concentration of NO at equilibrium - 1.0 - x is the concentration of Cl2 at equilibrium
03

Write the equilibrium constant expression

Use the balanced chemical equation to write the equilibrium constant expression: \[K = \frac{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
04

Substitute equilibrium concentrations into the equilibrium constant expression

Substitute the equilibrium concentrations from the ICE table into the equilibrium constant expression: \[1.6 \times 10^{-5} = \frac{(2.0 - 2x)^2 (1.0 - x)}{x^2}\]
05

Solve for x

Now, we have a quadratic equation that we need to solve for x: \[1.6 \times 10^{-5} = \frac{(4 - 8x + 4x^2) (1.0 - x)}{x^2}\] Solve for x using the quadratic formula and neglect the negative value because concentrations cannot be negative. After solving for x, we get: \[x \approx 7.98 \times 10^{-3}\]
06

Calculate equilibrium concentrations

Now that we have the value of x, we can find the equilibrium concentrations: - Equilibrium concentration of NOCl: \(x\approx 7.98\times 10^{-3}M\) - Equilibrium concentration of NO: \(2.0-2x \approx 1.984M\) - Equilibrium concentration of Cl2: \(1.0-x \approx 0.992M\) Thus, the equilibrium concentrations of all species are: - \([\mathrm{NOCl}] \approx 7.98\times 10^{-3}M\) - \([\mathrm{NO}] \approx 1.984M\) - \([\mathrm{Cl}_{2}] \approx 0.992M\)

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Most popular questions from this chapter

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) c. \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) d. \(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\)

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calcu- late the value of \(K\) for the reaction.

At a particular temperature a \(2.00-\mathrm{L}\) flask at equilibrium contains \(2.80 \times 10^{-4}\) mole of \(\mathrm{N}_{2}, 2.50 \times 10^{-5}\) mole of \(\mathrm{O}_{2}\), and \(2.00 \times 10^{-2}\) mole of \(\mathrm{N}_{2} \mathrm{O}\). Calculate \(K\) at this temperature for the reaction $$ 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2} \mathrm{O}(g) $$ If \(\left[\mathrm{N}_{2}\right]=2.00 \times 10^{-4} M,\left[\mathrm{~N}_{2} \mathrm{O}\right]=0.200 M\), and \(\left[\underline{ \left.\mathrm{O}_{2}\right]}=\right.\) \(0.00245 M\), does this represent a system at equilibrium?

The equilibrium constant is \(0.0900\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. \(P_{H, O}=1.00 \mathrm{~atm}, P_{\mathrm{CL}, \mathrm{O}}=1.00 \mathrm{~atm}, P_{\mathrm{HOC}}=1.00 \mathrm{~atm}\) b. \(P_{\mathrm{H}_{2} \mathrm{O}}=200\). torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=49.8\) torr, \(P_{\mathrm{Ho} \mathrm{C}}=21.0\) torr c. \(P_{\mathrm{H}_{0} \mathrm{O}}=296\) torr, \(P_{\mathrm{C}_{6} \mathrm{O}}=15.0\) torr, \(P_{\mathrm{HO} \mathrm{C}}=20.0\) torr
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