QuestionApril 24, 2025

Consider the following reaction. 2NO_(2)(g)leftharpoons N_(2)O_(4)(g) K_(p)=3.10at298K Calculate K_(c) at this temperature. square

Consider the following reaction. 2NO_(2)(g)leftharpoons N_(2)O_(4)(g) K_(p)=3.10at298K Calculate K_(c) at this temperature. square
Consider the following reaction. 2NO_(2)(g)leftharpoons N_(2)O_(4)(g) K_(p)=3.10at298K Calculate K_(c) at this temperature. square

Solution
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Answer

K_c = 75.07 Explanation 1. Use the relationship between K_p and K_c The formula is K_p = K_c(RT)^{\Delta n}, where \Delta n is the change in moles of gas. Rearrange to find K_c: K_c = \frac{K_p}{(RT)^{\Delta n}}. 2. Calculate \Delta n For the reaction 2NO_2(g) \rightleftharpoons N_2O_4(g), \Delta n = 1 - 2 = -1. 3. Substitute values Use R = 0.0821 \, L \cdot atm / mol \cdot K, T = 298 \, K, and K_p = 3.10. K_c = \frac{3.10}{(0.0821 \cdot 298)^{-1}}. 4. Simplify the expression (0.0821 \cdot 298) = 24.4758, so: K_c = 3.10 \cdot 24.4758 = 75.07.

Explanation

1. Use the relationship between $K_p$ and $K_c$<br /> The formula is $K_p = K_c(RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas. Rearrange to find $K_c$: <br />$$K_c = \frac{K_p}{(RT)^{\Delta n}}.$$<br /><br />2. Calculate $\Delta n$<br /> For the reaction $2NO_2(g) \rightleftharpoons N_2O_4(g)$, $\Delta n = 1 - 2 = -1$.<br /><br />3. Substitute values<br /> Use $R = 0.0821 \, L \cdot atm / mol \cdot K$, $T = 298 \, K$, and $K_p = 3.10$. <br />$$K_c = \frac{3.10}{(0.0821 \cdot 298)^{-1}}.$$<br /><br />4. Simplify the expression<br /> $(0.0821 \cdot 298) = 24.4758$, so:<br />$$K_c = 3.10 \cdot 24.4758 = 75.07.$$
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