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Chapter 1: Problem 37

For a reaction involving nitrogen monoxide inside a sealed flask, the value for the reaction quotient \((Q)\) was found to be \(1.1 \times 10^{2}\) at a given point. If, after this point, the amount of NO gas in the flask increased, which reaction is most likely taking place in the flask? (A) \(\operatorname{NOBr}(g) \rightarrow \operatorname{NO}(g)+1 / \operatorname{Br}_{2}(g) \quad K_{\mathrm{C}}=3.4 \times 10^{-2}\) (B) \(2 \mathrm{NOCl}(g) \mapsto 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K_{\mathrm{c}}=1.6 \times 10^{-5}\) (C) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{\mathrm{c}}=4.0 \times 10^{6}\) (D) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g) \quad K_{\mathrm{c}}=4.2 \times 10^{2}\)

Short Answer

Expert verified
The most likely reaction taking place in the flask is (D): \(N_{2}(g) + O_{2}(g) \rightarrow 2 NO(g)\) with an equilibrium constant \(K_{\mathrm{c}} = 4.2 \times 10^{2}\).

Step by step solution

01

Compare Reaction Quotient and Equilibrium Constant

Take note that when \(Q > K_c\), the reaction shifts to the left, moving toward reactants. If \(Q < K_c\), the reaction shifts to the right, favoring the formation of products. Since the value of \(Q = 1.1 \times 10^{2}\) and the NO gas is increasing, the reaction shifts to the right.
02

Check Each Reaction Option

For the reaction to shift to the right, \(Q\) must be less than \(K_c\). So check each option and see which of the reactions have a \(K_c\) value greater than \(1.1 \times 10^{2}\). (A) \(K_c = 3.4 \times 10^{-2}\), (B) \(K_c = 1.6 \times 10^{-5}\) , (C) \(K_c = 4.0 \times 10^{6}\), (D) \(K_c = 4.2 \times 10^{2}\). From this, it is clear that only reactions (C) and (D) have \(K_c\) values higher than the \(Q\) value.
03

Identify Reactions where Nitrogen Monoxide Gas Increases

Since we know that the concentration of nitrogen monoxide gas is increasing, the reaction involved must be producing more nitrogen monoxide. This means that nitrogen monoxide must be a product in the reaction. Considering reactions (C) and (D), it is only in reaction (D) that nitrogen monoxide is among the products.

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Most popular questions from this chapter

A 0.1-molar solution of which of the following acids will be the best conductor of electricity? (A) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) (B) \(\mathrm{H}_{2} \mathrm{S}\) (C) \(\mathrm{HF}\) (D) \(\mathrm{HNO}_{3}\)

\(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) If the reaction above took place at standard temperature and pressure and 150 grams of \(\mathrm{CaCO}_{3}(\mathrm{s})\) were consumed, what was the volume of \(\mathrm{CO}_{2}(g)\) produced at STP? (A) 11 L (B) 22 L (C) 34 L (D) 45 L

Use the following information to answer questions 14-16 The radius of atoms and ions is typically measured in Angstroms \((A),\) which is equivalent to \(1 * 10^{-10} \mathrm{m} .\) Below is a table of information for three different elements. TABLE NOT AVAILABLE The phosphorus ion is larger than a neutral phosphorus atom, yet a zinc ion is smaller than a neutral zinc atom. Which of the following statements best explains why? (A) The zinc atom has more protons than the phosphorus atom. (B) The phosphorus atom is paramagnetic, but the zinc atom is diamagnetic. (C) Phosphorus gains electrons when forming an ion, but zinc loses them. (D) The valence electrons in zinc are further from the nucleus than those in phosphorus.

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ If, instead of copper, a nickel bar were to be used, could nickel be plated onto the zinc penny effectively? Why or why not? (A) Yes, nickel’s SRP is greater than that of zinc, which is all that is required for nickel to be reduced at the cathode (B) Yes, nickel is able to take electrons from the \(\mathrm{H}^{+}\) ions in solution, allowing it to be reduced (C) No, nickel's SRP is lower than that of \(\mathrm{H}^{+}\) ions, which means the only product being produced at the cathode would be hydrogen gas (D) No, nickel's SRP is negative, meaning it cannot be reduced in an electrolytic cell

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ When the cell is connected, which of the following reactions takes place at the anode? (A) \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) (B) \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}+2 e^{-}\) (C) \(2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_{2}(g)\) (D) \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}+2 e^{-}\)
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