Please use the options given to me **Question: 10 of 17**Consider the equilibrium system described by the chemical reaction below. A 1.00 L reaction vessel was filled with 2.00 mol SO₂ and 2.00 mol NO₂ and allowed to react at a high temperature. At equilibrium, there were 1.30 mol of NO in the vessel. Determine the concentrations of all reactants and products at equilibrium and then calculate the value of \( K_c \) for this reaction.**Chemical Reaction:** \[ \text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{NO}(g) \]**Instructions:** Based on the setup of your ICE table, construct the expression for \( K_c \) and then evaluate it. Do not combine or simplify terms.\[ K_c = \frac{\text{[product]} \times \text{[product]}}{\text{[reactant]} \times \text{[reactant]}} = \]**Choices for Inputs:** \[ [0] \quad [x] \quad [1.00] \quad [2.00] \quad [1.30] \quad [2.30] \quad [2.70] \quad [0.70] \] \[ [1.35] \quad [2.00 - x] \quad [2.00 + x] \quad [x]^2 \quad [2.00 - x]^2 \quad [2.00 + x]^2 \] \[ 1.9 \quad 0.54 \quad 0.29 \quad 3.4 \]**RESET Button:** A button labeled "RESET" is available for restarting the calculation process. ### Equilibrium Concentration Calculation Using ICE Table**Problem Description:**Consider the equilibrium system described by the chemical reaction below. A 1.00 L reaction vessel was filled with 2.00 mol of SO₂ and 2.00 mol of NO₂ and allowed to react at a high temperature. At equilibrium, there were 1.30 mol of NO in the vessel. Determine the concentrations of all reactants and products at equilibrium and then calculate the value of \( K_c \) for this reaction.\[ \text{SO}_2(g) + \text{NO}_2(g) \rightleftharpoons \text{SO}_3(g) + \text{NO}(g) \]**Steps to Solve:**1. **Initial Concentrations:** - \[\text{SO}_2(g)\]: 2.00 M - \[\text{NO}_2(g)\]: 2.00 M - \[\text{SO}_3(g)\]: 0.00 M - \[\text{NO}(g)\]: 0.00 M2. **Change in Concentrations (let change be x):** - \[\text{SO}_2(g)\]: -x - \[\text{NO}_2(g)\]: -x - \[\text{SO}_3(g)\]: +x - \[\text{NO}(g)\]: +x3. **Equilibrium Concentrations:** - \[\text{SO}_2(g)\]: \( 2.00 - x \) M - \[\text{NO}_2(g)\]: \( 2.00 - x \) M - \[\text{SO}_3(g)\]: \( x \) M - \[\text{NO}(g)\]: \( x \) M**Given:**At equilibrium, \(\text{NO} = 1.30\) M, hence \(x = 1.30\).4. **Resulting Equilibrium Values:** - \[\text{SO}_2(g)\]: \( 2.00 - 1.30 = 0.70 \) M - \[\text{NO}_2(g)\]: \( 2.00 - 1.30 = 0.70 \) M

For a chemical reaction, the equilibrium constant can be defined as the ratio between the amount of reactant and the amount of product which is used to determine chemical behavior.Write the reaction. SO2(g) + NO2(g)⇔SO3(g) + NO(g)At t=0 2 mol 2 mol 0 0At eqb. 2 mol-x 2 mol-x x x Write the expression for Keq. Keq=[NO][SO3][SO2][NO2] There are 1.3 mol of NO at equilibrium. Therefore, the value of x is 1.3. Substitute 1.3 mol for x to calculate Keq. Keq=[1.3 mol][1.3 mol][2-1.3 mol][2-1.3 mol] =3.44 Therefore, the value for Keq is 3.44.