Using the information below, we know that the rate law for this reaction is rate = k[A]2[B]2[C].
The following data were collected for the reaction A + B + C D + E:
To obtain the rate law, we must compare trials and determine how changes in the initial concentrations of the reactants affect the initial rate of the reaction. Comparing Trials 1 and 4, we see that [A] increased by a factor of 2 and the reaction rate increased by a factor of 4 = 22; thus, the reaction is second order with respect to A. Eliminate Rate = k[A][B][C] and Rate = k[A][B]2[C]2. Comparing Trials 1 and 3, we see that [B] increased by a factor of 2 and the reaction rate increased by a factor of 4 = 22; thus, the reaction is also second order with respect to B. By process of elimination you can choose Rate = k[A]2[B]2[C] as the correct answer without analyzing the rate law with respect to C. Doing so, however, we find that by comparing Trials 1 and 2, we see that [C] increased by a factor of 2 and so did the reaction rate; thus, the reaction is first order with respect to C. Therefore, the rate law for this reaction is given by Rate = k[A]2[B]2[C].
In the diagram, 1, 2, 3, and 4 represent, respectively:
Question 17 Answer Choices
A. Ea (forward), the activated complex, ΔH, Ea (reverse)
B. the activated complex, ΔH, Ea (forward), Ea (reverse)
C. the activated complex, Ea (forward), ΔH, Ea (reverse)
D. Ea (reverse), the activated complex, ΔH, Ea (forward)
In the diagram, 1, 2, 3, and 4 represent, respectively the activated complex, Ea (forward), ΔH, Ea (reverse).
Point 1 is the energy of the activated complex. Line 3 is the difference in energy between the products and reactants; this is ΔH. Thus, the answer must be "the activated complex, Ea (forward), ΔH, Ea (reverse)".
The formation constant (Kf) for the complex ion [Fe(en)3]2+ is 5.0 × 109, whereas the Kf for [Fe(ox)3]4- is 1.7 × 105 (en = 1,2 diaminoethane, ox = oxalate anion). Based on this information, 1,2-diaminoethane is a stronger ligand with iron(II) than the oxalate ion.
Formation constants are equilibrium constants for the formation of complex ions. The larger the equilibrium constant of a reaction, the larger the ratio of products to reactants at equilibrium will be. The oxalate complex with iron(II) has a lower Kf than the diamine complex. Therefore, any competitive reaction will favor the diamine complex, so [Fe(en)3]2+ is favored to form, not [Fe(ox)3]4-. Since both complexes have iron(II) as the central metal, there isn't any information to support the relationship suggested between oxidation state and the value of the formation constant. When comparing competing ligands, the one with the larger Kf will yield the more stable complex, which is due to the strongest Lewis acid/Lewis base interactions. Therefore, the diamine must be the stronger ligand. Both of the ligands in the question are bidentate, so there is no information to use for comparison between ligand type and value of Kf.
A sealed container contains NO2, a brownish-red gas, and N2O4, a colorless gas, at equilibrium at 0°C according to the following reaction: . The gas mixture will become lighter in color if the container is placed in dry ice and acetone at -78°C.
The system is at equilibrium at 0°C, so ΔG must equal 0. We also know that during the forward reaction 2 moles of gas are converted to 1 mole of gas, which results in a decrease in entropy (ΔS < 0). From this, the sign of ΔH can be determined because ΔG = 0 = ΔH - TΔS. Solving for ΔH gives us: ΔH = TΔS. ΔH must be negative for the forward reaction because ΔS is also negative. Since the forward reaction is exothermic, it will be thermodynamically favored by a decrease in temperature. As temperature falls, the reaction shifts to the right. N2O4 (colorless) increases and NO2 (brownish-red) decreases, leading to an overall lighter color.
The following reaction is allowed to reach equilibrium:
Cl2O(g) + 2 OH-(aq) <--> 2 OCl-(aq) + H2O(l)
The effect of then doubling the hydroxide concentration will be that the reaction quotient will be one-fourth the equilibrium constant, favoring the forward reaction.
First, eliminate "the reaction quotient will be four times the equilibrium constant, favoring the reverse reaction"; adding a reactant favors the forward reaction. Since the equilibrium constant, K, and the reaction quotient, Q, are inversely proportional to [OH-]2, doubling [OH-] will cause K to decrease by a factor of 22 = 4. Thus, the reaction quotient, Q, will be one-fourth of K, favoring the forward reaction.