For the reaction $\mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{IBr}(g), K_{c}=310\( at \)140^{\circ} \mathrm{C}$. Suppose that \(1.00 \mathrm{~mol}\) IBr in a \(5.00-\mathrm{L}\) flask is allowed to reach equilibrium at \(140^{\circ} \mathrm{C}\). What are the equilibrium concentrations of \(\mathrm{IBr}, \mathrm{I}_{2},\) and \(\mathrm{Br}_{2}\) ?

: Write the balanced chemical equation and make a table to keep track of the initial concentrations (I), the change in concentrations (C), and the equilibrium concentrations (E). The balanced chemical equation is given as: \[\mathrm{I}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2\operatorname{IBr}(g)\] Create the ICE table: | | \(\boldsymbol{\mathrm{I}_{2}}\)| \(\boldsymbol{\mathrm{Br}_{2}}\) | \(\operatorname{\boldsymbol{IBr}}\) | |----------|------------------|------------------|------------------| | Initial | 0 | 0 | 1 | | Change | \(+\)x | \(+\)x | \(-\)2x | | Equil. | x | x | 1\(-\)2x |

: Using the balanced equation and the equilibrium concentrations from the ICE table, write the equilibrium expression, set it equal to the given equilibrium constant, and substitute the equilibrium concentrations. Equilibrium expression: \[K_{c} = \frac{[\operatorname{IBr}]^{2}}{[\mathrm{I}_{2}][\mathrm{Br}_{2}]}\] Substituting equilibrium concentrations and given \(K_c\): \[310 = \frac{(1 - 2x)^{2}}{x^2}\]

: First, simplify the expression by squaring (1-2x): \[310 = \frac{1-4x+4x^{2}}{x^2}\] Then, cross-multiply to solve for the quadratic equation: \[310x^{2} = 1 - 4x + 4x^{2}\] Next, set everything to one side: \[0 = 4x^{2} - 310x^{2} - 4x + 1\] Combine the terms, and we get: \[0 = -306x^{2} - 4x + 1\]

: We use the quadratic formula: \[x =\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here, \(a = -306\), \(b = -4\), and \(c = 1\). Plug in the values: \[x =\frac{4 \pm \sqrt{(-4)^2 - 4(-306)(1)}}{2(-306)}\] Solving for x, we get two possible solutions: \(x = 0.0207\) or \(x = 0.0158\). Since 1-2x (concentration of IBr) must be positive, the solution \(0.0158\) will result in a negative equilibrium concentration for IBr. Therefore, we discard that solution. Thus, \(x = 0.0207\) is our best answer for the equilibrium concentrations.

: Now we can plug in the value of x we found back into our ICE table equilibrium concentrations: \([\mathrm{I}_{2}] = 0.0207~\mathrm{M}\) \([\mathrm{Br}_{2}] = 0.0207~\mathrm{M}\) \([\operatorname{IBr}] = 1 - 2(0.0207) = 0.958~\mathrm{M}\) The equilibrium concentrations at \(140^{\circ} \mathrm{C}\) are: \(\mathrm{I}_{2} = 0.0207~\mathrm{M}\) \(\mathrm{Br}_{2} = 0.0207~\mathrm{M}\) \(\operatorname{IBr} = 0.958~\mathrm{M}\)