Can you please solve question 6 and show all of the steps to the solution **Text for Educational Website**6. A mixture of 0.500 mol H₂ and 0.500 mol I₂ was placed in a 1.00-L stainless steel flask at 430°C. The equilibrium constant Kₑ for the reaction,\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]is 54.3 at this temperature. Calculate the concentrations of H₂, I₂, and HI at equilibrium.7. The pH of a 0.10 M solution of acetic acid (CH₃COOH) is 2.39. What is the \( K_a \) of acetic acid?

Given, the equilibrium chemical reaction is as follows,H2(g)+I2(g)⇌2HI(g)the equilibrium constant Kc =54.3 initial number of moles of H2, I2 are 0.500 moles each.volume of flask =1.00 Lwe are asked to calculate the equilibrium concentration of H2, I2 and HIThe number of moles of H2 = 0.500 mol The number of moles of I2 = 0.500 mol Volume of flask = 1.00L Now, concentration of H2 = no. of moles of H2 / volume of flask = 0.500 mol/ 1.00L = 0.500 mol concentration of I2 = no. of moles of I2 / volume of flask = 0.500 mol/ 1.00L = 0.500 mol the equilibrium chemical reaction is as follows, H2 (g)+I2(g)⇌2HI(g) Initial concentration 0.500 M 0.500M 0 M Change in concentration -x m -x m +2x m Equilibrium conc (0.500-x)M (0.500-x)m 2x m The equilibrium constant is given by, Kc = [HI]2 / [H2] [I2] = (2x)2 / (0.500-x)*(0.500-x) = (2x)2 / (0.500-x)2 Kc = (2x /0.500-x)2 2x /0.500-x = (Kc)0.5 =(54.3) 0.5 =7.37 2x /0.500-x = 7.372x= 7.37(0.500 - x) 2x= 3.685 – 7.37x 9.37x = 3.685 X= 0.393 M So, the equilibrium concentration of H2, I2 and HI are [H2] = (0.500 - x)= (0.500 – 0.393) M = 0.107M [I2] = (0.500 - x)= (0.500 – 0.393) M = 0.107M [HI] = (2x)= (2 * 0.393) M = 0.786M Therefore, the equilibrium concentration of H2, I2 and HI are 0.107 M, 0.107 M, 0.786 MTherefore, the equilibrium concentration of H2, I2 and HI are 0.107 M, 0.107 M, 0.786 M