Question
Step 1: \( 2NO\rightleftharpoons (NO)_2\) (fast)
Step 2: \((NO)_2+O_2\rightleftharpoons 2(NO)_2\) (slow)
The elementary steps in a proposed mechanism for the reaction \(2NO(g)+O_2(g)→2NO_2(g)\) are represented by the equations above. Which of the following is the rate law for the overall reaction that is consistent with the proposed mechanism?
A rate=\(k[NO]_2\)
B rate=\(k[NO][O_2]\)
C rate=\(k[NO]^2[O_2]\)
D rate=\(k[(NO)_2][O_2]\)
▶️Answer/Explanation
Ans:C
The rate of the forward reaction in step 1 depends on the collision of two NO molecules, so its rate law is \(rate_{forward}=k_f[NO]^2\). The rate of the reverse reaction in step 1 depends only on the concentration of \((NO)_2\), so its rate law is \(rate_{reverse}=k_r[(NO)_2]\). Because step 1 is an equilibrium, the forward and reverse rates are equal, meaning that \(k_f[NO]^2=k_r[(NO)_2]\). Step 2 is the slow (rate-determining) step, which depends on the collision between \((NO)_2\) and \(O_2\). Its rate law is \(rate_{Step 2}=k_2[(NO)_2][O_2]\). From Step 1, \(k_f[NO]^2=k_r[(NO)_2]\), therefore \([(NO)_2]=\frac{k_f}{k_r}[NO]^2\). Substituting this expression into the rate law for step 2 yields the equation \(rate_{Step 2}=(k2)(\frac{k_f}{k_r})[NO]^2[O_2] = k'[NO]^2[O_2]\). If the rate constant for the overall reaction k′ is renamed as k, the final rate law expression is rate=\(k[NO]^2[O_2]\).
Question
\( H_2(g)+I_2(g)→2HI(g)\)
For the reaction between \(H_2 ~and~ I_2\) , shown above, the following two-step reaction mechanism is proposed.
Step 1: \(I_2\rightleftharpoons 2I\) (fast equilibrium)
Step 2: \(H_2+2I→2HI\) (slow)
What is the rate law expression for this reaction if the second step is rate determining?
A Rate=\(k[I_2]\)
B Rate=\(k[H_2][I]^2\)
C Rate=\(k[H_2][I_2]\)
D Rate=\(k[I_2][I]^2\)
▶️Answer/Explanation
Ans:C
The rate law of a reaction is set by the stoichiometry of the rate-determining step. The rate law for the rate-determining step (step 2) is rate\(=k_2[H_2][I]^2\). However, intermediates should not be included in the rate law. Step 1 is a fast equilibrium reaction, and the rates of the forward and reverse reactions are equal: \(k_f[I_2]=k_r[I]^2\). Solving for \([I]_2\) gives [I]^2=\farc{k_f[I_2]}{k_r}. Substituting for \([I]^2\) in the rate law for step 2 gives rate=\frac{k_2[H_2]k_f[I_2]}{k_r}, which is equal to rate=\(k[H_2][I_2]\).
Question
\( H_2(g)+I_2(g)→2HI(g)\)
For the reaction between \(H_2 ~and~ I_2\) , shown above, the following two-step reaction mechanism is proposed.
Step 1: \(I_2\rightleftharpoons 2I\) (fast equilibrium)
Step 2: \(H_2+2I→2HI\) (slow)
What is the rate law expression for this reaction if the second step is rate determining?
A Rate=\(k[I_2]\)
B Rate=\(k[H_2][I]^2\)
C Rate=\(k[H_2][I_2]\)
D Rate=\(k[I_2][I]^2\)
▶️Answer/Explanation
Ans:C
The rate law of a reaction is set by the stoichiometry of the rate-determining step. The rate law for the rate-determining step (step 2) is rate\(=k_2[H_2][I]^2\). However, intermediates should not be included in the rate law. Step 1 is a fast equilibrium reaction, and the rates of the forward and reverse reactions are equal: \(k_f[I_2]=k_r[I]^2\). Solving for \([I]_2\) gives [I]^2=\farc{k_f[I_2]}{k_r}. Substituting for \([I]^2\) in the rate law for step 2 gives rate=\frac{k_2[H_2]k_f[I_2]}{k_r}, which is equal to rate=\(k[H_2][I_2]\).
Question
Step 1: 2X(g)⇄X2(g) (fast)
Step 2: \(X_2(g)+Y(g)→X_2Y(g)\) (slow)
The rate law for the hypothetical reaction \(2X(g)+Y(g)→X_2Y(g)\) is consistent with the mechanism shown above. Which of the following mathematical equations provides a rate law that is consistent with this mechanism?
A rate\(=k[X]^2\)
B rate\(=k[X_2][Y]\)
C rate\(=k[X]^2[Y]\)
D rate\(=k[2X][Y]\)
▶️Answer/Explanation
Ans: C
Since step 2 is the rate-determining step, rate\(=k_2[X_2][Y]\) . X2 is an intermediate formed in a fast equilibrium step, thus, \(k_{forward}[X]^2=kreverse[X_2]\). Rearranging to solve for \([X_2]\) and substituting in the first expression, rate\(=k[X]^2[Y]\), where \(k=k_2\frac{k_{forward}}{k_{reverse}}\).
Question
Reaction A: \(4 HCl(g) + O_{2}(g) \rightleftharpoons 2Cl_{2}(g) + 2 H_{2}O(g)\)
Reaction B:\( N_{2}O_{4}(g)\rightleftharpoons 2 NO_{2}\)(g)\)
Reaction C:\( H_{2}(g) + I_{2}(g)\rightleftharpoons 2 HI(g)\)
Reaction D: \(2 NH_{3}(g) \rightleftharpoons N_{2}(g) +3 H_{2}(g)\)
The reactions represented above are carried out in sealed, rigid containers and allowed to reach equilibrium. If the volume of each container is reduced from 1.0 L to 0.5 L at constant temperature, for which of the reactions will the amount of product(s) be increased?
(A) Reaction A
(B) Reaction B
(C) Reaction C
(D) Reaction D
▶️Answer/Explanation
Ans:B
Question
\(Ag^{+}(aq)+NH_{3}(aq)\rightleftharpoons Ag(NH_{3}^{+})(aq)\) \(K_{eq1}=2.0=2.0\times10^{3}\)
\(Ag(NH_{3})^{+}(aq)+NH_{3}(aq)\rightleftharpoons Ag(NH_{3}^{+})(aq) \) \(K_{eq2}=2.0=2.0\times10^{3}\)
Equal volumes of 0.1 M \(AgNO_{3}(aq)\) and 2.0 M\( NH_{3}\)(aq) are mixed and the reactions represented above occur. Which Ag species will have the highest concentration in the equilibrium system shown below, and why?
\(Ag^{+}(aq) + 2 NH_{3}(aq) \rightleftharpoons Ag(NH_{3})2^{+}\)(aq) \(K_{egs}\) =?
(A) Ag (aq), because \(K_{eq3}= 4\)
(B) Ag^{+}(aq), because \(K_{eq1}< K_{eq2}\)
(C) \(Ag(NH_{3})2(aq)\), because \(K_{eq3} = 1.6 \times 10^7\)
(D) \(Ag(NH_{3})2(aq)\), because \(K_{eq1}< K_{Ag}\)
▶️Answer/Explanation
Ans:A