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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

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$. If 2.0 moles of \(\mathrm{CO}_{2}\) is initially placed into a 5.0 - \(\mathrm{L}\) vessel, calculate the equilibrium concentrations of all species.

Nitrogen gas \(\left(\mathrm{N}_{2}\right)\) reacts with hydrogen gas \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) At \(200^{\circ} \mathrm{C}\) in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 atm of hydrogen gas. At equilibrium,the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the \(K_{\mathrm{p}}\) value for this reaction.

Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium:$$\mathrm{CoCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)$$ Purple Pink.What color will such an indicator be if rain is imminent?

For the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$.\(K_{\mathrm{p}}=3.5 \times 10^{4}\) at \(1495 \mathrm{K} .\) What is the value of \(K_{\mathrm{p}}\) for the following reactions at \(1495 \mathrm{K} ?\) a. \(\operatorname{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\). b. \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) c. \(\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g)\).

The reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\). If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of NO at equilibrium.
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