During a chemical reaction, NO(g) gets reduced and no nitrogen- containing compound is oxidized. Which of the following is a possible product of this reaction? (A) \(\mathrm{NO}_{2}(g)\) (B) \(\mathrm{N}_{2}(g)\) (C) \(\mathrm{NO}_{3}^{-}(a q)\) (D) \(\mathrm{NO}_{2}^{-}(a q)\)

Short Answer

Expert verified
The possible product of this reaction is \( N_{2}(g) \).

Step by step solution

01

Understand the concept of reduction

Reduction is a chemical reaction where an atom, ion, or molecule gains electrons. In other words, the oxidation state of the molecule decreases.
02

Consider the molecule being reduced

In this case, the molecule being reduced is NO(g). This means that the nitrogen-containing compound will gain electrons. Note that gaining electrons means the oxidation number decrease.
03

Identify which products are possible

In reduction reactions, the molecule gains electrons, meaning the oxidation state decreases. Thus, we need to find a molecule where nitrogen has a lower oxidation number than it does in NO(g). With an oxidation number of +2 in NO, nitrogen could decrease its oxidation number by forming either N2(g) or NO2^-(aq). N2 is a neutral molecule in which nitrogen has an oxidation number of 0, and NO2^- is an ion in which nitrogen has the oxidation number of +3. The other options (NO2(g) and NO3^-(aq)) show an increase in the oxidation number of nitrogen (+4 and +5 respectively), which contradicts the fact that NO is getting reduced.
04

Select the correct answer

Since we have determined that reduction requires a decrease in oxidation number, only N2(g) and NO2^-(aq) qualify, making them possible products. However, since no nitrogen- containing compound is oxidized, it indicates that there are no other nitrogen-containing compounds other than NO(g) in the reaction. Therefore, the production of NO2^-(aq) cannot occur because it would require a nitrogen-containing compound to be oxidized. Thus, the correct answer is N2(g).

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Most popular questions from this chapter

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ How many moles of electrons must be transferred to create 127 g of copper? (A) 1 mole of electrons (B) 2 moles of electrons (C) 3 moles of electrons (D) 4 moles of electrons

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. What is the approximate amount of heat released during the reaction? \(\begin{array}{ll}{\text { (A) }} & {1.5 \mathrm{kJ}} \\ {\text { (B) }} & {2.9 \mathrm{kJ}} \\ {\text { (C) }} & {5.9 \mathrm{kJ}} \\ {\text { (D) }} & {11.8 \mathrm{kJ}}\end{array}\)

Which of the following could be added to an aqueous solution of weak acid HF to increase the percent dissociation? (A) \(\operatorname{NaF}(s)\) (B) \(\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{NaOH}(\mathrm{s})\) (D) \(\mathrm{NH}_{3}(a q)\)

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$

Questions 45-48 refer to the following. Inside a calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) hydrocyanic acid (HCN), a weak acid, and 100.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) sodium hydroxide are mixed. The temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(28.5^{\circ} \mathrm{C}\) . The specific heat of the mixture is approximately \(4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C},\) and the density is identical to that of water. As \(\Delta T\) increases, what happens to the equilibrium constant and why? (A) The equilibrium constant increases because more products are created. (B) The equilibrium constant increases because the rate of the forward reaction increases. (C) The equilibrium constant decreases because the equilibrium shifts to the left. (D) The value for the equilibrium constant is unaffected by temperature and will not change.

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