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]1.2 \times 10^{-3}[/tex]
Answer (2)
Write the expression for the equilibrium constant K e q : K e q = [ H F ] 2 [ H 2 ] [ F 2 ] .
Substitute the given equilibrium concentrations into the K e q expression: K e q = ( 5.82 × 1 0 − 2 ) 2 ( 8.4 × 1 0 − 3 ) ( 8.4 × 1 0 − 3 ) .
Calculate the value of K e q : K e q = 0.020831119141247738 .
Express the calculated K e q value in scientific notation: K e q = 2.1 × 1 0 − 2 .
Explanation
Setting up the problem First, we need to calculate the equilibrium constant, K e q , for the given reaction. The reaction is: 2 H F ( g ) ⟷ H 2 ( g ) + F 2 ( g ) The equilibrium constant expression is given by: K e q = [ H F ] 2 [ H 2 ] [ F 2 ] We are given the 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
Calculating Keq Now, we substitute the given concentrations into the K e q expression: 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
Expressing Keq in scientific notation Finally, we express the calculated K e q value 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 Therefore, the value of K e q for the reaction is approximately 2.1 × 1 0 − 2 .
Examples
Understanding chemical equilibrium is crucial in many real-world applications. For instance, in the Haber-Bosch process, which synthesizes ammonia ( N H 3 ) from nitrogen and hydrogen, controlling the equilibrium conditions (temperature, pressure, and concentrations) is essential to maximize ammonia production. Similarly, in the production of various chemicals and pharmaceuticals, manipulating equilibrium constants allows chemists to optimize reaction yields and minimize waste, leading to more efficient and sustainable processes. By understanding and applying the principles of chemical equilibrium, we can design and control chemical reactions to meet specific needs in industry, medicine, and environmental science.
The equilibrium constant for the reaction 2 H F ( g ) ⟷ H 2 ( g ) + F 2 ( g ) was calculated using the concentrations provided. Substituting the concentrations into the expression K e q = [ H F ] 2 [ H 2 ] [ F 2 ] yielded a value of approximately 2.1 × 1 0 − 2 . Therefore, the answer is option A: 2.1 × 1 0 − 2 .
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