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Chapter 16: Problem 67

Consider the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}_{\text {thombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$For each of the following mixtures of reactants and products at \(25^{\circ} \mathrm{C},\) predict the direction in which the reaction will shift to reach equilibrium. a. \(P_{\mathrm{NO}_{2}}=P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.0 \mathrm{atm}\) b. \(P_{\mathrm{NO}_{2}}=0.21 \mathrm{atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.50 \mathrm{atm}\) c. \(P_{\mathrm{NO}_{2}}=0.29 \mathrm{atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm}\)

Short Answer

Expert verified
a. No shift in either direction, the reaction is already at equilibrium. b. The reaction will shift to the left, favoring the formation of NO2. c. The reaction will shift to the left, favoring the formation of NO2.

Step by step solution

01

a. Determine the direction for Reaction 1 with given pressures.

Here, the initial pressures are \(P_{NO_2} = P_{N_2O_4} = 1.0\;atm\). According to the balanced equation for Reaction 1, two moles of NO2 are being converted into one mole of N2O4. We can calculate the reaction quotient (Q) for Reaction 1: Reaction 1: \(Q=[N_2O_4]/[NO_2]^2\) Substitute the given pressures: \(Q=(1.0)/(1.0)^2=1\) Compare Q and K_c (K_c is also 1 for this Reaction at 25°C). Since \(Q=K_c\), the reaction is already at equilibrium, and there will be no shift in either direction.
02

b. Determine the direction for Reaction 1 with given pressures.

Here, the initial pressures are \(P_{NO_2} = 0.21\;atm\) and \(P_{N_2O_4} = 0.50\;atm\). Calculate the reaction quotient (Q) for Reaction 1: Substitute the given pressures: \(Q=(0.50)/(0.21)^2=11.30\) Compare Q and K_c. Since \(Q>K_c\), the reaction will shift to the left, favoring the formation of NO2.
03

c. Determine the direction for Reaction 1 with given pressures.

Here, the initial pressures are \(P_{NO_2} = 0.29\;atm\) and \(P_{N_2O_4} = 1.6\;atm\). Calculate the reaction quotient (Q) for Reaction 1: Substitute the given pressures: \(Q=(1.6)/(0.29)^2=18.97\) Compare Q and K_c. Since \(Q>K_c\), the reaction will shift to the left, favoring the formation of NO2. From the given exercise, only Reaction 1 had partial pressures to analyze. Thus, the directions for Reaction 1 have been predicted based on the partial pressures provided.

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Most popular questions from this chapter

For the reaction at \(298 \mathrm{K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ}\) and \(-176.6 \mathrm{J} / \mathrm{K},\) respectively. What is the value of \(\Delta G^{\circ}\) at 298 K? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {surr }}\) ? \(\Delta S ?\) \(\Delta S_{\text {univ }}\) ? Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

Consider the following reaction at \(298 \mathrm{K}:\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ An equilibrium mixture contains \(\mathrm{O}_{2}(g)\) and \(\mathrm{SO}_{3}(g)\) at partial pressures of 0.50 atm and 2.0 atm, respectively. Using data from Appendix \(4,\) determine the equilibrium partial pressure of \(\mathrm{SO}_{2}\) in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(g)\) at 1.00 atm and \(\mathrm{Br}_{2}(g)\) at 1.00 atm were mixed in a 1.00-L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ},\) and \(\Delta S^{\circ} .\)

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)
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