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Chapter 13: Problem 9

The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.

Short Answer

Expert verified
The value of the equilibrium constant (K) depends on the temperature of the system (option 'c') and the nature of the reactants and products (option 'd'). It is independent of the initial concentrations of reactants and products (options 'a' and 'b'). Changes in temperature can affect the balance between forward and reverse reaction rates, while the specific chemical properties of reactants and products determine the unique equilibrium constant value for each reaction.

Step by step solution

01

Option a: The initial concentrations of the reactants

The value of the equilibrium constant (K) is independent of the initial concentrations of the reactants. Equilibrium conditions depend on a fixed ratio of the concentrations of reactants and products, irrespective of their initial concentrations. Hence, option 'a' is incorrect.
02

Option b: The initial concentrations of the products

Similar to option 'a', the value of the equilibrium constant (K) is independent of the initial concentrations of the products. The equilibrium concentrations are determined by the fixed ratio between the concentrations of reactants and products, which does not depend on their initial concentrations. So, option 'b' is also incorrect.
03

Option c: The temperature of the system

The value of the equilibrium constant (K) is dependent on the temperature of the system for a given reaction. A change in temperature affects the equilibrium constant because it influences the balance between the forward and reverse reaction rates. If a reaction is endothermic, an increase in temperature will shift the equilibrium to favor the products, increasing the value of K. Conversely, if a reaction is exothermic, an increase in temperature will shift the equilibrium to favor the reactants, decreasing the value of K. Therefore, option 'c' is correct.
04

Option d: The nature of the reactants and products

The value of the equilibrium constant (K) depends on the nature of the reactants and products involved in the reaction. The equilibrium constant represents the balance between the forward and reverse reactions, which in turn depends on the specific chemical properties of the reactants and products. Each reaction has a unique equilibrium constant value that is specific to the reactants and products involved. So, option 'd' is correct. In conclusion, the value of the equilibrium constant (K) depends on the temperature of the system (option 'c') and the nature of the reactants and products (option 'd').

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Most popular questions from this chapter

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?

Consider the following reaction: $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2},\) and \(\mathrm{CO}_{2}\) are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \(^{14} \mathrm{C}\) will \(^{14} \mathrm{C}\) be found only in \(\mathrm{CO}\) molecules for an indefinite period of time? Explain.

For the reaction $$\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s)$$ \(K=400\) . at \(35.0^{\circ} \mathrm{C} .\) If 2.00 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and $\mathrm{NH}_{4} \mathrm{HS}\( are placed in a \)5.00-\mathrm{L}$ vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium?

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. $2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$ b. $\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)$ c. $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)$ d. $4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)$

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\) , equilibrium is reached when 50.0\(\%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?
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