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Chapter 13: Problem 53

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ What is the partial pressure of \(\mathrm{NO}\) in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of \(0.80\) and \(0.20\) atm, respectively?

Short Answer

Expert verified
The partial pressure of \(\mathrm{NO}\) in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is approximately \(0.204\, \text{atm}\).

Step by step solution

01

Write the reaction and equilibrium expression

Write down the given reaction and the equilibrium expression for Kp: \(\mathrm{N}_{2} (\mathrm{g}) + \mathrm{O}_{2} (\mathrm{g}) \rightleftharpoons 2 \mathrm{NO} (\mathrm{g})\) \( K_{\mathrm{p}} = \frac{\text{[NO]}^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} \)
02

Define the changes in partial pressures

Since we are given the initial pressures of N2 and O2, let's define the changes in their partial pressures with \(x\): Change in N2 pressure: -x Change in O2 pressure: -x Change in NO pressure: +2x
03

Write the expressions for the equilibrium partial pressures

Incorporate the changes in partial pressures into the equilibrium expression of each species: \([\mathrm{N}_2]_{eq} = 0.80-x\) \([\mathrm{O}_2]_{eq} = 0.20-x\) \([\mathrm{NO}]_{eq} = 2x\)
04

Substitute the equilibrium expressions into the Kp equation

Substitute the expressions for the equilibrium partial pressures into the Kp equation given: \( 0.050 = \frac{(2x)^2}{(0.80-x)(0.20-x)} \)
05

Solve for x

Solve the equation above for x: Solving the equation algebraically or numerically, we get: \(x \approx 0.102\)
06

Calculate equilibrium partial pressure of NO

Finally, calculate the equilibrium partial pressure of NO using the value of x: \([\mathrm{NO}]_{eq} = 2x = 2(0.102) \approx 0.204\, \text{atm}\) The partial pressure of NO in equilibrium with N2 and O2 is approximately 0.204 atm.

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Most popular questions from this chapter

A \(1.604-g\) sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and \(6.400 \mathrm{~g}\) oxygen gas are sealed into a 2.50-L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is \(0.326 \mathrm{~atm}\), and the pressure of water vapor is \(4.45\) atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

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?

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. \(2.0\) moles of pure \(\mathrm{NOCl}\) in a \(2.0-\mathrm{L}\) flask b. \(1.0\) mole of \(\mathrm{NOCl}\) and \(1.0 \mathrm{~mole}\) of \(\mathrm{NO}\) in a \(1.0-\mathrm{L}\) flask c. \(2.0\) moles of \(\mathrm{NOCl}\) and \(1.0 \mathrm{~mole}\) of \(\mathrm{Cl}_{2}\) in a \(1.0-\mathrm{L}\) flask

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is \(93.71 \%\) carbon by mass, and a \(0.256\) -mole sample of naphthalene has a mass of \(32.8 \mathrm{~g}\). What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene \((s) \rightleftharpoons\) naphthalene \((g)\) $$ K=4.29 \times 10^{-6}(\text { at } 298 \mathrm{~K}) $$ If \(3.00 \mathrm{~g}\) solid naphthalene is placed into an enclosed space with a volume of \(5.00 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\), what percentage of the naphthalene will have sublimed once equilibrium has been established?

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)
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