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3)
The production of ammonia occurs according to the following reversible reaction:
$\mathrm{N}_{2}+3 \mathrm{H}_{2}=2 \mathrm{NH}_{3}$
An equilibrium mixture for the reaction at $500 \mathrm{~K}$ has the following composition:
\begin{tabular}{|c|c|}
\hline Component & \begin{tabular}{c}
Equilibrium \\
concentration $\left(\mathrm{kmol} \mathrm{m}^{-3}\right)$
\end{tabular} \\
\hline $\mathrm{N}_{2}$ & $3.0 \times 10^{-2}$ \\
\hline $\mathrm{H}_{2}$ & $3.7 \times 10^{-2}$ \\
\hline $\mathrm{NH}_{3}$ & $1.6 \times 10^{-2}$ \\
\hline
\end{tabular}
Calculate the equilibrium constants for the forward and reverse reactions, and show how they are related.
$\left[1.7 \times 10^{2}, 5.9 \times 10^{-3}\right]$
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Answer from Sia

Posted 5 months ago

Solution

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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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