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Chapter 12: Problem 14

Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

Short Answer

Expert verified
At the given temperature, with an equilibrium constant of \(1.7 \times 10^{-8}\), there will be a large amount of unreacted starting material when the reaction reaches equilibrium. This reaction is not likely to be a good source of products at this temperature, as the small K value indicates that the equilibrium favors the reactants. To improve product yield, a different temperature or catalyst may be necessary.

Step by step solution

01

Analyze the given equilibrium constant (K) value

The equilibrium constant (K) for this reaction is \(1.7 \times 10^{-8}\). This is a very small value, which indicates that at equilibrium, the concentration of reactants is much higher than the concentration of products.
02

Determine the amount of unreacted starting material

Due to the K value being very small, it suggests that at equilibrium, there is a large amount of unreacted starting material. Consequently, only a small fraction of starting material is converted into products.
03

Evaluate the reaction as a source of products at this temperature

Since there is a significant amount of unreacted starting material at equilibrium, we can conclude that the reaction at this particular temperature is not likely to be a good source of products. The small K value illustrates that the reaction favors the reactants, making it difficult to obtain a high yield of products from this reaction. To improve the yield of products, a different temperature or possibly a catalyst may need to be explored to shift the equilibrium in favor of product formation.

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Most popular questions from this chapter

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

Explain the difference between \(K, K_{\mathrm{p}},\) and \(Q\).

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) A friend asks the following: "I know we have been told that if a mixture of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) is at equilibrium and more of \(\mathrm{A}\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more \(\mathrm{B} ?^{\prime \prime}\) What do you tell your friend?

Consider the following exothermic reaction at equilibrium: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$. Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

For the reaction:$$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 \mathrm{M},\) calculate the equilibrium \(\mathrm{O}_{3}\) concentration.
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