Question

(a) How many moles of HI are produced when \(2.00 \mathrm{~mol}\) \(\mathrm{H}_{2}\) and \(2.00 \mathrm{~mol} \mathrm{} \mathrm{I}_{2}\) are reacted at \(700 \mathrm{~K}\) ? (Reaction is \(79 \%\) complete.) (b) Addition of \(0.27 \mathrm{~mol} \mathrm{} \mathrm{I}_{2}\) to the system increases the yield of HI to \(85 \%\). How many moles of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and HI are now present? (c) From the data in part (a), calculate \(K_{\text {eq }}\) for the reaction at \(700 \mathrm{~K}\).

Short answer

Originally forms 3.16 mol HI; with added I_2 forms 3.40 mol HI, 0 H_2, 0.27 I_2; K_c = 14.10

Step-by-step solution

- Stoichiometric Calculation for Initial Reaction

Determine the number of moles of HI that would be produced if the reaction went to completion. The balanced equation is: \[ \text{H}_2 + \text{I}_2 \rightarrow 2\text{HI} \] Given: 2.00 mol \text{H}_2 and 2.00 mol \text{I}_2. The reaction has a 1:1 molar ratio of reactants, hence the limiting reactant will dictate the amount of product formed. If the reaction completed: \[ 2.00 \text{~mol H}_2 + 2.00 \text{~mol I}_2 \rightarrow 2(2.00) = 4.00 \text{~mol HI} \]

- Account for Reaction Completion

The reaction is only 79% complete. Calculate the number of moles of HI produced: \[ 4.00 \text{~mol HI} \times 0.79 = 3.16 \text{~mol HI} \]

- Addition of Iodine to Increase HI Yield

With the addition of 0.27 mol \(I_2\), the new yield of HI is 85%. First, redo the stoichiometry to find out how many moles of \(\text{HI}\) are formed with the added iodine. The initial moles of \(I_2\) is now: \[ 2.00 \text{~mol I}_2 + 0.27 \text{~mol I}_2 = 2.27 \text{~mol I}_2 \] Assume the amount will remain in the 1:1 ratio with \(H_2\), so only 2.00 mol \(I_2\) will be completely reacted due to limiting reactant \(\text{H}_2\) \[ 2(2.00) = 4.00 \text{~mol HI}_{85\text{%}} = 4.00 \text{~mol HI} \times 0.85 = 3.40 \text{~mol HI} \]

- Remaining Reactants

With 0.27 mol I_2 added, the initial 2 mol of H_2 partly contribute to HI production. Remaining moles: \[ 2.00 \text{~mol ~ H}_2 -2 = 0.00 \text{~mol H}_2 \] and of I_2 used is 2.27: \[ 2.27 \text{~mol I}_2-2= 0.27 \text{~mol}\]So: \( 0.00 \text{~mol H}_2, 0.27 \text{~mol I}_2, 3.40 \text{~mol HI} \)

- Calculating Equilibrium Constant \(K_{eq}\)

For the reaction to calculate \(K_{eq}\) at 700 K, at state of 3.16 HI moles originally nearly complete at equilibrium: Using initial moles: 2 mol &2; Find concentration M: [H2]=[I2]=initial -changeUsing ICE method calculate [H2]=[I2] is \(0.42 \text{~mol/L}\) &HI is up to \(1.58 M\) plugging: \[K_c = \frac{[HI]}{[H2][I2]} = \text{(1.58^2)/(0.42 \text{x} 0.42)=14.1} \]

Key concepts

reaction stoichiometry

Chemical reactions follow specific ratios, which is defined by the balanced chemical equation. This is known as reaction stoichiometry. For the reaction \(\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}\), understanding stoichiometry allows us to predict how much of each product will be formed given the amounts of reactants. Here, one mole of \(\text{H}_2\) reacts with one mole of \(\text{I}_2\) to produce two moles of \(\text{HI}\). If you start with 2.00 moles of \(\text{H}_2\) and 2.00 moles of \(\text{I}_2\), and the reaction goes to completion, you can ideally produce 4.00 moles of \(\text{HI}\). This ratio is critically important when calculating expected yields of products.

reaction completeness

Not all reactions proceed to full completion. In real-world scenarios, many reactions reach a state where only a certain percentage of the reactants are converted into products. This concept is referred to as reaction completeness. For instance, in part (a) of the problem, the reaction completeness is 79%. This means only 79% of the theoretical yield is achieved. Using the theoretical yield of 4.00 moles \(\text{HI}\) from complete reaction stoichiometry, we can calculate the actual yield: \[ 4.00 \text{~mol HI} \times 0.79 = 3.16 \text{~mol HI} \] Only 3.16 moles of \(\text{HI}\) are practically produced as opposed to the ideal 4.00 moles. Understanding the concept of reaction completeness is vital in predicting the actual outcomes of chemical processes.

limiting reactant

A limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. For the reaction \(\text{H}_2 + \text{I}_2 \rightarrow 2\text{HI}\), both \(\text{H}_2\) and \(\text{I}_2\) start with equal moles (2.00 each). When additional 0.27 moles of \(\text{I}_2\) are added for part (b), accounting for improved yield rates (85%), \(\text{I}_2\) might initially appear to exceed \(\text{H}_2\) in amount. However, \(\text{H}_2\) remains the limiting reactant with its initial count dictating the potential moles of \(\text{HI}\) formed. With 2.00 moles of \(\text{H}_2\) present, it reacts with an equal 2.00 moles of \(\text{I}_2\), leaving the extra \(\text{I}_2\) unreacted and 0.27 moles of \(\text{I}_2\) free in mixture post-reaction.

equilibrium constant

The equilibrium constant \(\text{K}_{eq}\) is an unchanging value for a particular reaction at a given temperature. It's a measure of the ratio of products to reactants when the reaction is at equilibrium. From part (c) of the problem, the reaction at 700K yields the equation: \(\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}\). At equilibrium, using the initial moles and calculated yields, we find the molar concentrations: \([H2] = [I2] = 0.42 \text{~} mol/L\) and \([HI]=1.58 \text{~} mol/L\). Therefore, the equilibrium constant \(\text{K}_{eq}\) can be calculated as: \[ K_c = \frac{[HI]^2}{[H2][I2]} = \frac{1.58^2}{0.42 \times 0.42} = 14.1 \] This value signifies the ratio of concentrations of products to reactants at the equilibrium state.