Answers are given, please work through and explain **Equilibrium Reaction Study****Reaction Details:**Consider the chemical reaction:\[ \text{CO}_{(g)} + \text{H}_2_{(g)} \leftrightarrow \text{C}_{(s)} + \text{H}_2\text{O}_{(g)} \]with an equilibrium constant (\( K_c \)) of 17.9 at 125.0°C.**Initial Conditions:**Assume you start with a 500.0 mL flask containing:- 0.5000 moles of carbon (\( \text{C}_{(s)} \))- 1.500 moles of water vapor (\( \text{H}_2\text{O}_{(g)} \))- 0.4630 moles of carbon monoxide (\( \text{CO}_{(g)} \))- 0.2280 moles of hydrogen gas (\( \text{H}_2_{(g)} \))**Tasks:**a. Calculate the reaction quotient for this equilibrium process.b. Determine the concentrations of all species at equilibrium.c. Find the \( K_p \) for this reaction at 125.0°C.**Answers:**- Reaction quotient: 7- Concentrations at equilibrium: - Carbon monoxide: 3 - Hydrogen gas: 0.7 - Water vapor: 0.2

Given reaction: At 125 oC, Kc = 17.9 The volume of the flask = 500 mL = 0.5 L Moles of C(s) = 0.5 moles moles of water vapour = 1.500 moles moles of CO = 0.4630 moles moles of H2 gas = 0.2280 molesCalculating the concentrations of all the species: [H2O] = molesvolume(L)=1.500 mol0.5 L = 3 M[CO] = 0.4630 mol0.5 L = 0.926 M[H2] = 0.2280 mol0.5 L = 0.456 M Please note that the concentration of solid species is taken as 1, as it does not change. So, [C]=1. Reaction quotient, Q measures the relative amounts of products and reactants present during a reaction at a particular point in time. So, let's calculate Q at the start of the reaction. Q = [H2O][CO] [H2]Substituting the concentration values:Q = 30.926 × 0.456Q = 7Because Q < K, the reaction will move in the forward direction. CO(g) + H2(g) → C(s) + H2O(g) I 0.926 M 0.456 M - 3C -x -x - 3+xE 0.926-x 0.456-x - 3+xKc = 17.9Kc =[H2O][CO] [H2]17.9 = (3+x) (0.926-x)( 0.456-x)17.9 (x2 - 1.382x + 0.422) = 3 +x17.9x2 - 24.738x + 7.55 = 3 +x17.9x2 - 25.738x + 4.55 =0On solving this quadratic equation, we get:x = 1.23, x=0.206x cannot be greater than the original value, so 1.23 gets eliminated.We get, x = 0.206 Equilibrium concentrations:[CO] = 0.926-x = 0.926-0.206=0.72[H2] = 0.456-x = 0.456-0.206=0.25[H2O] = 3+x = 3+0.206 = 3.206Kp =Kc (RT)∆n where:R = gas constant = 0.0821 L atm K-1 mol-1T= temp= 125°C = 125+273 K =398 K ∆n = mol of gaseous products-mol of gaseous reactants = 1-(1+1)=1-2=-1Kp = 17.9 (0.0821×398)-1 Kp = 0.55