Part 0 Leaming Goal: To use the equilibrium constant and Le Chateliert principle to determine how a reaction will respond to extemal factors. A reaction is at equilibrium when the concentrations of the reactants and products no tonger change over time. This does not mean the reaction is over; rather, two competing reactions continue to occur sitmuittaneousty at equat rates. The two competing feactions are the forward reaction (reactants \( \rightarrow \) products) and the reverse reaction (products \( \rightarrow \) reactants). II a reaction at equilibrium is subjected to a stress, the concentrations of reactants and products adjust to reontabitith equiftortum. This ts catted te Chateliats prinotple. A stress might be a change in the concentration of reactants or products, a change in the volume of the reaction container, a change in tmperature, of the addition of a catatyst. The concentration of \( \mathrm{I}_{2}(\mathrm{~g}) \) is increased to 1.5 M , disrupting equilibrium. Catoulate the new ratio of products to reactants with this higher concentration of lodine. Assume that the reaction has not yet regathed equititoflum. Express your answer using two signifieant figures. View Avaliable Hint( ( ) \( \square \) Bubmit Previoun Answers

To solve this problem, we need to understand the equilibrium expression for the reaction involving iodine (\( \mathrm{I}_2 \)). The equilibrium constant (\( K \)) for a reaction is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their coefficients in the balanced equation. ### Step 1: Write the balanced chemical equation Assuming the reaction is: \[ \mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g) \] ### Step 2: Write the equilibrium expression The equilibrium constant expression for this reaction is: \[ K = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} \] ### Step 3: Identify the initial concentrations Let's denote: - The initial concentration of \( \mathrm{I}_2 \) before the increase is \( [\mathrm{I}_2]_{initial} \). - After the concentration of \( \mathrm{I}_2 \) is increased to 1.5 M, we have: \[ [\mathrm{I}_2] = 1.5 \, \text{M} \] ### Step 4: Determine the change in concentrations Since the reaction has not yet reached equilibrium after the increase in \( [\mathrm{I}_2] \), we can denote the change in concentration of \( \mathrm{I} \) as \( +2x \) (since 2 moles of \( \mathrm{I} \) are produced for every mole of \( \mathrm{I}_2 \) that reacts) and the change in concentration of \( \mathrm{I}_2 \) as \( -x \). ### Step 5: Set up the expression for the new concentrations At this point, we can express the concentrations as: - \( [\mathrm{I}_2] = 1.5 - x \) - \( [\mathrm{I}] = 2x \) ### Step 6: Substitute into the equilibrium expression Substituting these into the equilibrium expression gives: \[ K = \frac{(2x)^2}{(1.5 - x)} \] ### Step 7: Calculate the new ratio of products to reactants To find the new ratio of products to reactants, we need to express it as: \[ \text{Ratio} = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} = \frac{(2x)^2}{(1.5 - x)} \] However, we need the value of \( K \) to calculate the exact ratio. If \( K \) is known, we can solve for \( x \) and then find the ratio. ### Step 8: Final expression Since we do not have the value of \( K \), we can express the ratio in terms of \( x \): \[ \text{Ratio} = \frac{4x^2}{1.5 - x} \] If you have the value of \( K \), please provide it, and we can proceed to calculate the specific ratio. Otherwise, we can express the ratio in terms of \( x \) as shown above.