Questions
Which of the following best explains why the combustion reactions represented in the table are exothermic?
(A) The number of bonds in the reactant molecules is greater than the number of bonds in the product molecules.
(B) The number of bonds in the reactant molecules is less than the number of bonds in the product molecules.
(C) The energy required to break the bonds in the reactants is greater than the energy released in forming the bonds in the products.
(D) The energy required to break the bonds in the reactants is less than the energy released in forming the bonds in the products.
▶️Answer/Explanation
Ans: D
Combustion reactions are exothermic because the energy released in forming the new bonds in the products (CO2 and H2O) is greater than the energy required to break the bonds in the reactants (hydrocarbons and O2).
Therefore, the correct answer is (D) The energy required to break the bonds in the reactants is less than the energy released in forming the bonds in the products.
Question
\(2 NO_2(g) → N_2O_4\) (g)
dark brown colorless
The dimerization of \(NO _2\)(g), an exothermic process, is represented by the equation above.
A mixture of \(NO _2\)(g) and \(N_2O_4\)(g) is at equilibrium in a rigid reaction vessel. If the temperature of the mixture is decreased, then
(A)\( [NO_2]\) will increase and the mixture will turn a darker brown
(B)\( [N_2O_4]\) will increase and the mixture will turn a lighter brown
(C) \( [N_2O_4]\) will decrease and the mixture will turn a lighter brown
(D) no change will be observed
▶️Answer/Explanation
Ans:B
The correct answer is (B) \([N_2O_4]\) will increase and the mixture will turn a lighter brown.
The dimerization of \(NO_2\) to form \(N_2O_4\) is an exothermic process, which means it releases heat. According to Le Chatelier’s principle, if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset.
So, if the temperature of the mixture is decreased, the system will try to counteract this change by producing more heat. It can do this by shifting the equilibrium to the right, towards the exothermic direction. This means more \(N_2O_4\) will be formed, and since \(N_2O_4\) is colorless, the mixture will turn a lighter brown. Therefore, option (B) is correct.
Question
\(2 NO_2(g) → N_2O_4\) (g)
dark brown colorless
The dimerization of NO2(g), an exothermic process, is represented by the equation above.
The forward reaction is thermodynamically favored at which of the following temperatures?
(A) All temperatures
(B) Low temperatures only
(C) High temperatures only
(D) No temperature
▶️Answer/Explanation
Ans:B
The dimerization of \(NO_2(g)\) to form \(N_2O_4(g)\) is an exothermic process, meaning it releases heat. According to Le Chatelier’s principle, for an exothermic reaction, increasing the temperature will shift the equilibrium position towards the reactants, favoring the reverse reaction. Conversely, decreasing the temperature will favor the forward reaction, leading to the formation of more \(N_2O_4(g)\).
Therefore, the forward reaction is thermodynamically favored at low temperatures. Thus, the correct answer is:
(B) Low temperatures only
Questions
\(2H_{2}O_{2}(l)\rightarrow 2H_{2}O(l)+O_{2(g)}\)
The exothermic process represented above is best classified as a
(A) physical change because a new phase appears in the products
(B) physical change because O2(g) that was dissolved comes out of solution
(C) chemical change because entropy increases as the process proceeds
(D) chemical change because covalent bonds are broken and new covalent bonds are formed
▶️Answer/Explanation
Ans: D
The given reaction involves the decomposition of hydrogen peroxide (\(H_{2}O_{2}\)) into water (\(H_{2}O\)) and oxygen gas (\(O_{2}\)).
Option (A): Physical changes typically involve changes in state or appearance without altering the chemical composition of the substance. In this reaction, while a phase change does occur (from liquid to gas for \(O_{2}\)), the chemical composition of the substances changes, so it’s not purely a physical change.
Option (B): This option suggests that the appearance of \(O_{2}(g)\) is due to it coming out of solution. However, the appearance of \(O_{2}(g)\) is due to its formation as a product of a chemical reaction rather than coming out of a solution. So, this option is not accurate.
Option (C): Entropy increasing doesn’t necessarily classify a reaction as chemical or physical. While entropy often increases in chemical reactions, it’s not a definitive indicator of a chemical change.
Option (D): Chemical changes involve the breaking and forming of chemical bonds, resulting in the formation of new substances with different chemical properties. In this reaction, covalent bonds in \(H_{2}O_{2}\) are broken to form \(H_{2}O\) and \(O_{2}\), indicating a chemical change.
So, the best classification for the given reaction is:
\[
\text{(D)} \quad \text{chemical change because covalent bonds are broken and new covalent bonds are formed}
\]