# Solving EquilibriumDr. Dahm is attempting to store a sample of HI. Little does he realize that there exists an equilibrium between HI, I₂, and H₂. If he starts with an initial pressure of HI of 1.51 atm, H₂ of 0.45 atm and I₂ of 1.17 atm, find the pressures at equilibrium. The equilibrium constant (K) for this reaction is 1.34 x 10⁻². Calculate the pressure of I₂ and H₂ when the reaction comes to equilibrium.### Reaction\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]### Initial Pressures- \( \text{HI} = 1.51 \, \text{atm} \)- \( \text{H}_2 = 0.45 \, \text{atm} \)- \( \text{I}_2 = 1.17 \, \text{atm} \)### Change in Pressures - Change for \( \text{HI} = -2p \)- Change for \( \text{H}_2 = +p \)- Change for \( \text{I}_2 = +p \)### Equilibrium Pressures- \( \text{HI} = 1.51 - 2p \)- \( \text{H}_2 = 0.45 + p \)- \( \text{I}_2 = 1.17 + p \)### Calculations- Use the equilibrium constant \( K = 1.34 \times 10^{-2} \) to solve for the pressures:\[K = \frac{(0.45 + p)(1.17 + p)}{(1.51 - 2p)^2}\]Solve for \( p \) and substitute back to find the equilibrium pressures for \( \text{H}_2 \) and \( \text{I}_2 \).---### Additional ProblemDr. Dahm wants to make \( \text{N}_2\text{O} \) so he reacts \( \text{N}_2 \) with \( \text{O}_2 \) to give \( \text{N}_2\text{O} \). Unfortunately, this is an equilibrium process that has a K

Note- Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.” Here we to determine equilibrium pressure of H2 ,I2 and total equilibrium pressure in the following given equilibrium reaction.