QuestionMay 4, 2025

Suppose a 250. mL flask is filled with 1.4 mol of NO_(3) and 0.60 mol of NO_(2) The following reaction becomes possible: NO_(3)(g)+NO(g)leftharpoons 2NO_(2)(g) The equilibrium constant K for this reaction is 2.27 at the temperature of the flask. Calculate the equilibrium molarity of NO_(2) Round your answer to two decimal places.

Suppose a 250. mL flask is filled with 1.4 mol of NO_(3) and 0.60 mol of NO_(2) The following reaction becomes possible: NO_(3)(g)+NO(g)leftharpoons 2NO_(2)(g) The equilibrium constant K for this reaction is 2.27 at the temperature of the flask. Calculate the equilibrium molarity of NO_(2) Round your answer to two decimal places.
Suppose a 250. mL flask is filled with 1.4 mol of NO_(3) and 0.60 mol of NO_(2) The following reaction becomes possible: NO_(3)(g)+NO(g)leftharpoons 2NO_(2)(g) The equilibrium constant K for this reaction is 2.27 at the temperature of the flask. Calculate the equilibrium molarity of NO_(2) Round your answer to two decimal places.

Solution
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Answer

The equilibrium molarity of NO_2 is approximately 3.14 M. Explanation 1. Determine Initial Concentrations Calculate the initial concentrations of NO_3 and NO_2 using the formula \text{Concentration} = \frac{\text{moles}}{\text{volume}}. For NO_3, \text{Concentration} = \frac{1.4 \, \text{mol}}{0.250 \, \text{L}} = 5.6 \, \text{M}. For NO_2, \text{Concentration} = \frac{0.60 \, \text{mol}}{0.250 \, \text{L}} = 2.4 \, \text{M}. 2. Set Up the ICE Table Use an ICE (Initial, Change, Equilibrium) table to express the changes in concentration for each species. Let x be the change in molarity of NO_3 that reacts. The changes are: [NO_3] = 5.6 - x, [NO] = x, [NO_2] = 2.4 + 2x. 3. Write the Expression for K The equilibrium constant expression is K = \frac NO_2]^2} NO_3][NO . Substitute the expressions from the ICE table into this equation: K = \frac{(2.4 + 2x)^2}{(5.6 - x)(x)}. 4. Solve for x Substitute K = 2.27 into the equilibrium expression: 2.27 = \frac{(2.4 + 2x)^2}{(5.6 - x)(x)}. Solve this equation for x using algebraic methods or numerical approximation techniques. 5. Calculate Equilibrium Molarity of NO_2 Once x is found, calculate the equilibrium concentration of NO_2: [NO_2] = 2.4 + 2x. Round the result to two decimal places.

Explanation

1. Determine Initial Concentrations<br /> Calculate the initial concentrations of $NO_3$ and $NO_2$ using the formula $\text{Concentration} = \frac{\text{moles}}{\text{volume}}$. For $NO_3$, $\text{Concentration} = \frac{1.4 \, \text{mol}}{0.250 \, \text{L}} = 5.6 \, \text{M}$. For $NO_2$, $\text{Concentration} = \frac{0.60 \, \text{mol}}{0.250 \, \text{L}} = 2.4 \, \text{M}$.<br /><br />2. Set Up the ICE Table<br /> Use an ICE (Initial, Change, Equilibrium) table to express the changes in concentration for each species. Let $x$ be the change in molarity of $NO_3$ that reacts. The changes are: $[NO_3] = 5.6 - x$, $[NO] = x$, $[NO_2] = 2.4 + 2x$.<br /><br />3. Write the Expression for K<br /> The equilibrium constant expression is $K = \frac{[NO_2]^2}{[NO_3][NO]}$. Substitute the expressions from the ICE table into this equation: $K = \frac{(2.4 + 2x)^2}{(5.6 - x)(x)}$.<br /><br />4. Solve for x<br /> Substitute $K = 2.27$ into the equilibrium expression: $2.27 = \frac{(2.4 + 2x)^2}{(5.6 - x)(x)}$. Solve this equation for $x$ using algebraic methods or numerical approximation techniques.<br /><br />5. Calculate Equilibrium Molarity of $NO_2$<br /> Once $x$ is found, calculate the equilibrium concentration of $NO_2$: $[NO_2] = 2.4 + 2x$. Round the result to two decimal places.
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