Consider the reaction:
[tex]2 HF(g) \longleftrightarrow H_2(g)+F_2(g)[/tex]
At equilibrium at 600 K, the concentrations are as follows:
[tex]
\begin{array}{l}
{[HF]=5.82 \times 10^{-2} M} \\
{\left[H_2\right]=8.4 \times 10^{-3} M} \\
{\left[F_2\right]=8.4 \times 10^{-3} M}
\end{array}
[/tex]
What is the value of [tex]K _{ eq }[/tex] for the reaction expressed in scientific notation?
A. [tex]2.1 \times 10^{-2}[/tex]
B. [tex]2.1 \times 10^2[/tex]
C. [tex]2.2 \times 10^{-1}[/tex]
D. [tex]1.2 \times 10^{-3}[/tex]
Answer (2)
Calculate K e q using the formula: K e q = [ H F ] 2 [ H 2 ] [ F 2 ] .
Substitute the given concentrations: K e q = ( 5.82 × 1 0 − 2 ) 2 ( 8.4 × 1 0 − 3 ) ( 8.4 × 1 0 − 3 ) .
Calculate the value: K e q = 0.020831119141247738 .
Express in scientific notation and round to two significant figures: K e q = 2.1 × 1 0 − 2 .
Explanation
Problem Analysis The problem provides the equilibrium concentrations for the reaction 2 H F ( g ) ⟷ H 2 ( g ) + F 2 ( g ) at 600 K. We are asked to calculate the equilibrium constant, K e q , for this reaction and express it in scientific notation.
Defining the Equilibrium Constant The equilibrium constant, K e q , is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. For the given reaction, the expression for K e q is: K e q = [ H F ] 2 [ H 2 ] [ F 2 ] where [ H 2 ] , [ F 2 ] , and [ H F ] are the equilibrium concentrations of hydrogen, fluorine, and hydrogen fluoride, respectively.
Substituting the Values We are given the following equilibrium concentrations: [ H F ] = 5.82 × 1 0 − 2 M [ H 2 ] = 8.4 × 1 0 − 3 M [ F 2 ] = 8.4 × 1 0 − 3 M Substituting these values into the expression for K e q , we get: K e q = ( 5.82 × 1 0 − 2 ) 2 ( 8.4 × 1 0 − 3 ) ( 8.4 × 1 0 − 3 ) K e q = 3.38724 × 1 0 − 3 7.056 × 1 0 − 5 K e q = 0.020831119141247738
Calculating Keq and Expressing in Scientific Notation Now, we express the value of K e q in scientific notation: K e q = 0.020831119141247738 = 2.0831119141247738 × 1 0 − 2 Rounding to two significant figures, we get: K e q = 2.1 × 1 0 − 2
Final Answer The value of K e q for the reaction is 2.1 × 1 0 − 2 .
Examples
The equilibrium constant is a crucial concept in chemistry, especially in industrial processes. For instance, in the Haber-Bosch process for ammonia synthesis ( N 2 + 3 H 2 ⇌ 2 N H 3 ), understanding and manipulating the equilibrium constant allows chemists to optimize reaction conditions (temperature, pressure, and reactant ratios) to maximize ammonia production. This optimization directly impacts fertilizer production, which is vital for global agriculture and food supply. By carefully adjusting these conditions based on the equilibrium constant, industries can achieve higher yields and greater efficiency, leading to significant economic and environmental benefits.
The equilibrium constant K e q for the reaction is calculated to be 2.1 × 1 0 − 2 , corresponding to option A. This was determined by substituting the equilibrium concentrations into the formula and calculating the result. Thus, the final answer is K e q = 2.1 × 1 0 − 2 .
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