Consider the equilibrium $X_2+Y_2\rightleftharpoons ?P$. Find the stoichiometric coefficient of $P$ using the data given in the following table.
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In equilibrium problems with tables, the Law of Mass Action is your primary tool. If the temperature doesn't change, \( K_c \) is your constant anchor.
Concept:
The equilibrium constant \( K_c \) for a reaction remains constant at a given temperature regardless of the individual equilibrium concentrations. For the reaction \( X_2 + Y_2 \rightleftharpoons nP \), the expression is:
\[
K_c = \frac{[P]^n}{[X_2][Y_2]}
\]
Step 1: Set up the ratio for two different sets of data.
Since \( K_c \) is constant, we can equate the ratios from both rows:
\[
\frac{(2.52 \times 10^{-2})^n}{(1.14 \times 10^{-2})(0.12 \times 10^{-2})} = \frac{(3.08 \times 10^{-2})^n}{(0.92 \times 10^{-2})(0.22 \times 10^{-2})}
\]
Step 2: Simplify the denominators.
• Denominator 1: \( 1.14 \times 0.12 \times 10^{-4} = 0.1368 \times 10^{-4} \)
• Denominator 2: \( 0.92 \times 0.22 \times 10^{-4} = 0.2024 \times 10^{-4} \)
Step 3: Solve for n.
Rearrange the equation:
\[
\left( \frac{3.08}{2.52} \right)^n = \frac{0.2024 \times 10^{-4}}{0.1368 \times 10^{-4}}
\]
\[
(1.222)^n \approx 1.479
\]
Since \( 1.222^2 \approx 1.49 \), we find that \( n = 2 \).
