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.
$$
\left[1.7 \times 10^{2}, 5.9 \times 10^{-3}\right]
$$

Question

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.
$$
\left[1.7 \times 10^{2}, 5.9 \times 10^{-3}\right]
$$

Accepted Answer

∻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 $ N_2 + 3H_2 \rightleftharpoons 2NH_3 $, the equilibrium constant $ K_c $ is given by $ 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 $ K_c $. Thus, $ K_c = \frac{(1.6 \times 10^{-2})^2}{(3.0 \times 10^{-2})(3.7 \times 10^{-2})^3} $. Calculating this gives $ 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, $ K_c' = \frac{1}{K_c} $. ‖ ‖ 4 ⋮ **Calculate the equilibrium constant for the reverse reaction (Kc')**: Using the value of $ K_c $ from step 2, $ 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. ⚹◊What is the expression for the equilibrium constant Kc for the forward reaction N2 + 3H2 ⇌ 2NH3?⍭ Generate me a similar question◊

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