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.

1. Write the equilibrium expression<br /> The equilibrium constant expression for the reaction is given by:<br />\[<br />K = \frac{[NO_2]^2}{[NO_3][NO]}<br />\]<br />where $[NO_2]$, $[NO_3]$, and $[NO]$ are the molar concentrations of $NO_2$, $NO_3$, and $NO$ at equilibrium, respectively.<br /><br />---<br /><br />2. Define initial concentrations<br /> The volume of the flask is 250 mL or 0.250 L. Using the formula for molarity ($M = \frac{\text{moles}}{\text{volume in liters}}$), calculate the initial concentrations:<br />\[<br />[NO_3]_0 = \frac{1.4 \, \text{mol}}{0.250 \, \text{L}} = 5.6 \, \text{M}<br />\]<br />\[<br />[NO_2]_0 = \frac{0.60 \, \text{mol}}{0.250 \, \text{L}} = 2.4 \, \text{M}<br />\]<br />Assume that initially, there is no $NO$ present:<br />\[<br />[NO]_0 = 0 \, \text{M}<br />\]<br /><br />---<br /><br />3. Set up changes in concentration using an ICE table<br /> Let $x$ represent the change in concentration of $NO$ formed at equilibrium. The changes in concentration for each species are as follows:<br />- $[NO_3]$ decreases by $x$: $[NO_3] = 5.6 - x$<br />- $[NO]$ increases by $x$: $[NO] = x$<br />- $[NO_2]$ increases by $2x$: $[NO_2] = 2.4 + 2x$<br /><br />At equilibrium, the concentrations are:<br />\[<br />[NO_3] = 5.6 - x, \quad [NO] = x, \quad [NO_2] = 2.4 + 2x<br />\]<br /><br />---<br /><br />4. Substitute into the equilibrium expression<br /> Substitute the equilibrium concentrations into the equilibrium constant expression:<br />\[<br />K = \frac{[NO_2]^2}{[NO_3][NO]}<br />\]<br />\[<br />2.27 = \frac{(2.4 + 2x)^2}{(5.6 - x)(x)}<br />\]<br /><br />---<br /><br />5. Solve for $x$<br /> Expand and rearrange the equation:<br />\[<br />2.27 = \frac{(2.4 + 2x)^2}{(5.6 - x)(x)}<br />\]<br />\[<br />2.27 (5.6x - x^2) = (2.4 + 2x)^2<br />\]<br />Expand both sides:<br />\[<br />2.27 (5.6x - x^2) = (2.4)^2 + 2(2.4)(2x) + (2x)^2<br />\]<br />\[<br />2.27 (5.6x - x^2) = 5.76 + 9.6x + 4x^2<br />\]<br />Distribute $2.27$ on the left-hand side:<br />\[<br />12.712x - 2.27x^2 = 5.76 + 9.6x + 4x^2<br />\]<br />Rearrange all terms to one side:<br />\[<br />0 = 5.76 - 3.112x + 6.27x^2<br />\]<br /><br />---<br /><br />6. Solve the quadratic equation<br /> Use the quadratic formula:<br />\[<br />ax^2 + bx + c = 0 \quad \Rightarrow \quad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}<br />\]<br />Here, $a = 6.27$, $b = -3.112$, and $c = -5.76$. Substituting these values:<br />\[<br />x = \frac{-(-3.112) \pm \sqrt{(-3.112)^2 - 4(6.27)(-5.76)}}{2(6.27)}<br />\]<br />\[<br />x = \frac{3.112 \pm \sqrt{9.686 + 144.4032}}{12.54}<br />\]<br />\[<br />x = \frac{3.112 \pm \sqrt{154.0892}}{12.54}<br />\]<br />\[<br />x = \frac{3.112 \pm 12.41}{12.54}<br />\]<br />Solve for the two possible values of $x$:<br />\[<br />x = \frac{3.112 + 12.41}{12.54} = \frac{15.522}{12.54} \approx 1.24<br />\]<br />\[<br />x = \frac{3.112 - 12.41}{12.54} = \frac{-9.298}{12.54} \approx -0.74<br />\]<br />Since $x$ must be positive, we take $x = 1.24$.<br /><br />---<br /><br />7. Calculate the equilibrium concentration of $NO_2$<br /> At equilibrium, $[NO_2] = 2.4 + 2x$. Substituting $x = 1.24$:<br />\[<br />[NO_2] = 2.4 + 2(1.24) = 2.4 + 2.48 = 4.88 \, \text{M}<br />\]<br /><br />---