Unsure how to proceed please help **Equilibrium Calculation for Chemical Reaction**Consider the equilibrium system described by the chemical reaction below. At equilibrium, a 1.5 L reaction vessel contained a mixture of 5.0 × 10⁻² M SO₃, 3.5 × 10⁻³ M O₂, and 3.0 × 10⁻³ M SO₂ at 800 K. What are the values of \( K_c \) and \( K_p \) for the reaction at this temperature?\[ 2 \text{ SO}_3(g) + \text{ O}_2(g) \rightleftharpoons 2 \text{ SO}_3(g) \]---**Task:**Based on the given data, set up the expression for \( K_c \) and then evaluate it. Do not combine or simplify terms.\[ K_c = \frac{{[SO_3]^2}}{{[SO_2]^2 \cdot [O_2]}} = \_\_\_\_ \]**Choices for entering the answer in the expression:**- \( [5.0 \times 10^{-2}] \)- \( [3.5 \times 10^{-3}] \)- \( [3.0 \times 10^{-3}] \)- \( [2.5 \times 10^{-8}] \)- \( [2.5 \times 10^{-1}] \)- \( [2.3 \times 10^{-1}] \)- \( [2.0 \times 10^{-9}] \)- \( [1.3 \times 10^{3}] \)- \( [3.3 \times 10^{-9}] \)- \( [2.1 \times 10^{-4}] \)- \( [4.8 \times 10^{9}] \)**RESET Button:**Use the reset if you need to clear your current selections.---**Instructions:**To solve this problem, follow these steps:1. Use the concentrations of SO₃, O₂, and SO₂ provided to write the expression for \( K_c \).2. Select appropriate values from the given options to complete the expression.3. Calculate the \( K_c \) value.This exercise helps in understanding how to determine equilibrium constants based on the concentration of reactants and products in a chemical reaction.

Given equilibrium equation is 2SO2g+O2g⇌2SO3g Concentration of the reactant and the product at equilibrium is given by, SO3=5.0×10−2 MO2=3.5×10−3 MSO2=3.0×10−3 MTemperature, T=800 Equilibrium constant of the given reaction is given by KC=SO32SO22O2 Substitute the equilibrium concentration in the above formula KC=SO32SO22O2KC=5.0×10−223.0×10−323.5×10−3KC=7.9×104 L mol−1 This is the required equilibrium constant. Kp for the given reaction is given by Kp=KCRTnf−niWhere,KC is equilibrium constatnR is gas consantnf and ni are the number of moles of gas on reactant and product.T is the temperature Substitute above R=0.0821 Latm mol−1K−1KC=7.937×104L mol−1T=800 Knf=2 and ni=3 It gives Kp=7.937×104L mol−10.0821 Latm mol−1K−1⋅800 K2−3Kp=1208.36 atm−1Kp=1.2×103 atm−1 This is the required Kp for the given reaction.Answer: Kc=7.9×104L mol−1Kp=1.2×103 atm−1