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Chapter 13: Problem 34

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}{\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)} \\ {P_{\mathrm{NH}_{3}}=3.1 \times 10^{-2} \mathrm{atm}} \\\ {P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm}} \\\ {P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}}\end{array}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If $P_{\mathrm{N}_{2}}=0.525 \mathrm{atm}, P_{\mathrm{NH}_{3}}=0.0167 \mathrm{atm},\( and \)P_{\mathrm{H}_{2}}=0.00761 \mathrm{atm},$ does this represent a system at equilibrium?

Short Answer

Expert verified
When using the given equilibrium pressures, the value of the equilibrium constant \(K_p\) is found to be 37.85. Using the new pressures, the calculated \(K_p'\) value is 33.92. The two values are not equal, which means the system is not at equilibrium with the new set of pressures.

Step by step solution

01

Write down the balanced equation and the expression for Kp

The balanced equation for the reaction is given as: \( N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \) We can write the expression for Kp as: \[ K_{p} = \frac{[NH_3]^2}{[N_2][H_2]^3} \] where [NH3], [N2], and [H2] represent the partial pressures of each gas.
02

Calculate the Kp using the given equilibrium pressures

Now, let's plug in the given equilibrium pressures into the Kp expression: \[ K_p = \frac{(3.1 \times 10^{-2})^2}{(8.5 \times 10^{-1})\times (3.1 \times 10^{-3})^3}= \frac{9.61 \times 10^{-4}}{2.5380125 \times 10^{-8}} \] Now, evaluate the equilibrium constant value: \[K_p= 37.85\]
03

Calculate the Kp using the new pressures

Now let's plug in the new pressures of N2, H2, and NH3 into the Kp expression and calculate the new Kp value: \[K_p' = \frac{(0.0167)^2}{(0.525)(0.00761)^3} = \frac{0.00027889}{8.22298 \times 10^{-6}}\] Evaluate the new Kp value: \[K_p'= 33.92\]
04

Compare the two Kp values to check for equilibrium

Let's compare the two Kp values that we have calculated. The initial Kp value was 37.85, and the new Kp value is 33.92. These values are not exactly the same, which suggests that the system is not at equilibrium with the new pressures provided.

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Most popular questions from this chapter

For the reaction $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),$ consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or \((b)\) you \(\operatorname{mix} 1.5\) moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

Consider the decomposition equilibrium for dinitrogen pentoxide: $$2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ At a certain temperature and a total pressure of 1.00 atm, the $\mathrm{N}_{2} \mathrm{O}_{5}\( is 0.50\)\%$ decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0,\) will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to 0.50\(\% ?\) Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of \(10.0 .\)

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 \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) 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 NOCl and 1.0 mole of NO in a 1.0 - flask c. 2.0 moles of \(\mathrm{NOCl}\) and 1.0 mole of \(\mathrm{Cl}_{2}\) in a \(1.0-\mathrm{L}\) flask

At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ At a low temperature, dry ice (solid \(\mathrm{CO}_{2} ),\) calcium oxide, and calcium carbonate are introduced into a \(50.0-\mathrm{L}\) reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C},\) resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2} .\) For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900^{\circ} \mathrm{C} ?\) a. \(655 \mathrm{g} \mathrm{CaCO}_{3}, 95.0 \mathrm{g}\) CaO, \(P_{\mathrm{CO}_{2}}=2.55 \mathrm{atm}\) b. $780 \mathrm{g} \mathrm{CaCO}_{3}, 1.00 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}$ c. $0.14 \mathrm{g} \mathrm{CaCO}_{3}, 5000 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}$ d. $715 \mathrm{g} \mathrm{CaCO}_{3}, 813 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{atm}$
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