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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

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For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),\) consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or (b) you mix 1.5 moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\) \(\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} \mathrm{M},\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=\) \(2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

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