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15240672380's Question
Chemistry
Posted 5 months ago

3)

The production of ammonia occurs according to the following reversible reaction:
N2+3H2=2NH3
\mathrm{N}_{2}+3 \mathrm{H}_{2}=2 \mathrm{NH}_{3}


An equilibrium mixture for the reaction at 500 K500 \mathrm{~K} has the following composition:
\begin{tabular}{|c|c|}
\hline Component & \begin{tabular}{c} 
Equilibrium \\
concentration (kmolm3)\left(\mathrm{kmol} \mathrm{m}^{-3}\right)
\end{tabular} \\
\hline N2\mathrm{N}_{2} & 3.0×1023.0 \times 10^{-2} \\
\hline H2\mathrm{H}_{2} & 3.7×1023.7 \times 10^{-2} \\
\hline NH3\mathrm{NH}_{3} & 1.6×1021.6 \times 10^{-2} \\
\hline
\end{tabular}

Calculate the equilibrium constants for the forward and reverse reactions, and show how they are related.
[1.7×102,5.9×103]
\left[1.7 \times 10^{2}, 5.9 \times 10^{-3}\right]
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Answer from Sia
Posted 5 months ago
Solution
1
Write the expression for the equilibrium constant (Kc) for the forward reaction: The equilibrium constant for the forward reaction can be expressed in terms of the concentrations of the products and reactants raised to the power of their stoichiometric coefficients. For the reaction N2+3H22NH3 N_2 + 3H_2 \rightleftharpoons 2NH_3 , the equilibrium constant Kc K_c is given by Kc=[NH3]2[N2][H2]3 K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}
2
Calculate the equilibrium constant for the forward reaction (Kc): Using the given equilibrium concentrations, we can substitute them into the expression for Kc K_c . Thus, Kc=(1.6×102)2(3.0×102)(3.7×102)3 K_c = \frac{(1.6 \times 10^{-2})^2}{(3.0 \times 10^{-2})(3.7 \times 10^{-2})^3} . Calculating this gives Kc=2.56×1041.5063×1041.7×102 K_c = \frac{2.56 \times 10^{-4}}{1.5063 \times 10^{-4}} \approx 1.7 \times 10^2
3
Write the expression for the equilibrium constant (Kc') for the reverse reaction: The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction, Kc=1Kc K_c' = \frac{1}{K_c}
4
Calculate the equilibrium constant for the reverse reaction (Kc'): Using the value of Kc K_c from step 2, Kc=11.7×1025.9×103 K_c' = \frac{1}{1.7 \times 10^2} \approx 5.9 \times 10^{-3}
Answer
[1.7 × 10^2, 5.9 × 10^-3]
Key Concept
The equilibrium constant for a reaction is a ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For the reverse reaction, the equilibrium constant is the inverse of the forward reaction's equilibrium constant.
Explanation
The calculated equilibrium constants for the forward and reverse reactions show the relationship between them, where the product of the two constants is equal to 1, demonstrating the reciprocal nature of these constants in a reversible chemical reaction.

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