Questions
Which of the following modifications will increase the rate of the reaction the most?
(A) Using 2.0 M \(CH_{3}COOH\)(aq) instead of 2.0 M HCl(aq)
(B) Cooling the HCl(aq) to a lower temperature than it was in the original experiment
(C) Reducing the volume of the reaction vessel
(D) Using eggshells that are more finely powdered than those used in the original experiment
▶️Answer/Explanation
Ans: C
Since reaction is exothermic, so lowering temperature will increase rate of reaction Hence, cooling the HCl (aq.) to a lower temperture than it was in original experiment will increase the rate of reaction most \(\rightarrow\) option (B)
Question
\(N_2O(g) + CO(g) \)→ \(N_2(g) + CO_2(g)\)
The rate of the reaction represented above increases significantly in the presence of Pd(s). Which of the following best explains this observation?
(A) Pd increases the activation energy of the reaction.
(B) Pd absorbs the heat produced in the reaction.
(C) One of the reactants binds on the surface of Pd, which introduces an alternative reaction pathway with a lower activation energy.
(D) One of the products binds on the surface of Pd, which increases the reaction rate by decreasing the concentration of products in the mixture.
▶️Answer/Explanation
Ans:C
The significant increase in the rate of the reaction in the presence of Pd(s) suggests that Pd acts as a catalyst. A catalyst provides an alternative reaction pathway with lower activation energy, allowing the reaction to occur more readily without being consumed in the process.
(A) Pd increases the activation energy of the reaction.
This statement contradicts the concept of catalysis. Catalysts typically lower the activation energy required for a reaction, making it easier for the reaction to proceed.
(B) Pd absorbs the heat produced in the reaction.
This statement does not explain the observed increase in the rate of the reaction. Heat absorption or removal may affect reaction equilibrium or kinetics, but it does not typically serve as the primary mechanism for catalysis.
(C) One of the reactants binds on the surface of Pd, which introduces an alternative reaction pathway with a lower activation energy.
This statement aligns with the concept of catalysis. Catalysts often provide an alternative reaction pathway by facilitating the adsorption of reactant molecules onto their surface, allowing the reaction to proceed through a lower activation energy pathway.
(D) One of the products binds on the surface of Pd, which increases the reaction rate by decreasing the concentration of products in the mixture.
While product inhibition is a phenomenon observed in some reactions, it does not explain the observed increase in the rate of the reaction in the presence of Pd(s).
Therefore, the most suitable explanation for the observed increase in the reaction rate is: (C) One of the reactants binds on the surface of Pd, which introduces an alternative reaction pathway with a lower activation energy.
Question
Which of the following shows the relationship between \(K_1\) and\( K_2\) in the reactions represented above?
(A) \(K_{2}\)= \((K_{1})^{2}\)
(B)\( K_{2}\)=\(\sqrt{K_{1}}\)
(C) \( K_{2}\)=\(\frac{1}{(K_{1})^{2}}\)
(D) \(K_{2}\)=\(\frac{1}{K_{1}}\)
▶️Answer/Explanation
Ans:C
To establish the relationship between \(K_1\) and \(K_2\) for the given reactions, we can use the principle of the equilibrium constant (\(K_{eq}\)).
For the first reaction:
\[\text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftarrows \text{SO}_3(g) \quad \text{(Equation 1)}\]
And for the second reaction:
\[2 \text{SO}_3(g) \rightleftarrows 2 \text{SO}_2(g) + \text{O}_2(g) \quad \text{(Equation 2)}\]
Given that \(K_1\) and \(K_2\) represent the equilibrium constants for these reactions, we can determine the relationship between them by considering the relationship between the equilibrium constants for the forward and reverse reactions.
For Equation 2, the equilibrium constant (\(K_2\)) is related to the equilibrium constant of the reverse reaction (Equation 1), which is \(K_{1}\).
Since Equation 2 is the reverse of Equation 1 and multiplied by 2, we have:
\( K_{2}\)=\(\frac{1}{(K_{1})^{2}}\)
Question
\[
\mathrm{Li}_3 \mathrm{~N}(s)+2 \mathrm{H}_2(g) \rightleftarrows \mathrm{LiNH}_2(s)+2 \mathrm{LiH}(s) \quad \Delta H^{\circ}=-192 \mathrm{~kJ} / \mathrm{mol}_{r x n}
\]
Because pure \(\mathrm{H}_2\) is a hazardous substance, safer and more cost effective techniques to store it as a solid for shipping purposes have been developed. One such method is the reaction represented above, which occurs at \(200^{\circ} \mathrm{C}\).
Which of the following is the most likely reason that the reaction occurs at a significant rate only if the temperature of the reaction mixture is greater than 200°C?
(A) The reaction is exothermic.
(B) ΔS° for the reaction is negative.
(C) The reaction has a high activation energy.
(D) ΔG° < 0 whenT < 200°C.
▶️Answer/Explanation
Ans:C
The temperature dependence of reaction rates is often explained by the Arrhenius equation, which states that the rate constant (\(k\)) of a reaction increases exponentially with temperature (\(T\)). This relationship suggests that increasing the temperature accelerates the reaction rate.
Given that the reaction occurs at a significant rate only if the temperature of the reaction mixture is greater than 200°C, we can infer that the reaction likely has a high activation energy. Activation energy is the energy barrier that must be overcome for a reaction to proceed. If the activation energy is high, a significant amount of thermal energy (high temperature) is required to provide the reactant molecules with enough kinetic energy to overcome this barrier and initiate the reaction.
So, the correct answer is:
(C) The reaction has a high activation energy.