This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Step 1: Write the balanced chemical equation and set up an ICE table for moles. The reaction is between hydrogen gas and iodine gas to form hydrogen iodide gas. H_2(g) + I_2(g) 2HI(g) Initial moles: H_2 = 8.0 mol, I_2 = 5.0 mol, HI = 0 mol. Let x be the moles of H_2 and I_2 that react. According to the stoichiometry, 2x moles of HI will be formed. | Species | Initial (mol) | Change (mol) | Equilibrium (mol) | | :------ | :------------ | :----------- | :---------------- | | H_2 | 8.0 | -x | 8.0 - x | | I_2 | 5.0 | -x | 5.0 - x | | HI | 0 | +2x | 2x | Step 2: Calculate the value of x using the given equilibrium moles of HI. At equilibrium, 9.06 mol of hydrogen iodide (HI) are formed. So, 2x = 9.06 mol. x = (9.06)/(2) = 4.53 mol Step 3: Calculate the equilibrium moles of all species. Equilibrium moles of H_2 = 8.0 - x = 8.0 - 4.53 = 3.47 mol Equilibrium moles of I_2 = 5.0 - x = 5.0 - 4.53 = 0.47 mol Equilibrium moles of HI = 9.06 mol Step 4: Calculate the equilibrium constant K_c. The expression for K_c is: K_c = [HI]^2[H_2][I_2] Let the volume of the closed vessel be V dm^3. The concentrations are moles divided by volume. [H_2] = (3.47)/(V) mol/dm^3 [I_2] = (0.47)/(V) mol/dm^3 [HI] = (9.06)/(V) mol/dm^3 Substitute these into the K_c expression: K_c = ((9.06)/(V))^2((3.47)/(V))((0.47)/(V)) = ((9.06)^2)/(V^2)((3.47)(0.47))/(V^2) The volume V^2 terms cancel out: K_c = ((9.06)^2)/((3.47)(0.47)) = (82.0836)/(1.6309) K_c ≈ 50.33 Rounding to three significant figures, K_c = 50.3. Step 5: Calculate the equilibrium constant K_p. The relationship between K_p and K_c is given by K_p = K_c(RT)^ n. n is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants. For the reaction H_2(g) + I_2(g) 2HI(g): Moles of gaseous products = 2 (from 2HI) Moles of gaseous reactants = 1 (from H_2) + 1 (from I_2) = 2 So, n = 2 - 2 = 0. Since n = 0, (RT)^ n = (RT)^0 = 1. Therefore, K_p = K_c × 1 = K_c. K_p = 50.33 Rounding to three significant figures, K_p = 50.3. 3 done, 2 left today. You're making progress.