Two reactions and their equilibrium constants are given. (Arrows are going back and forward) A+2B↽−−⇀2C.           K1= 2.33 2C↽−−⇀D.                  K2= 0.114   Calculate the value of the equilibrium constant for the reaction D↽−−⇀A+2B. K=?

Suppose a 250. mL flask is filled with 0.70 mol of NO₂, 0.20 mol of NO and 1.9 mol of CO₂. The following reaction becomes possible: NO₂ (g) + CO (g) NO(g) + CO₂(g) The equilibrium constant K for this reaction is 0.715 at the temperature of the flask. Calculate the equilibrium molarity of NO₂. Round your answer to two decimal places.

Suppose a 500.mL flask is filled with 1.5mol of CO, 1.1mol of H2O and 1.6mol of H2. The following reaction becomes possible: CO(g)+H2O(g)+CO2(g)+H2(g)   The equilibrium constant K for this reaction is 0.840 at the temperature of the flask. Calculate the equilibrium molarity of CO2. Round your answer to two decimal places.

Calculating an equilibrium constant from a partial equilibrium composition Steam reforming of methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 75.0 L tank with 35. mol of methane gas and 16. mol of water vapor, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be 43. mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits. K = [] с x10 X Ś ? olo 18 Ar

Suppose a 250. mL flask is filled with 0.60 mol of SO₂ and 1.0 mol of SO3. This reaction becomes possible: 2SO₂(g) + O₂(g) — 2SO3(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of O₂. You can leave out the M symbol for molarity. initial change equilibrium SO₂ 0₂ 0 X 0 SO₂ 0 010 X Ś

Hydrogen and chlorine react to form hydrogen chloride, like this: H₂(g) + Cl₂(g) - 2 HCl(g) Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, chlorine, and hydrogen chloride has the following composition: compound concentration at equilibrium H₂ 0.26M Cl₂ HCI Kc Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits. = 1.4M 0 0.97 M 0 x10 X

Consider the following equilibrium. N2(s) + O2(g) 2 NO(g) At 2300 K, the equilibrium constant K - 1.7 x 10. Suppose that 0.015 mol NO(g), 0.16 mol N, (9), and 0.16 mol O, (9) are placed into a 10.0 L flask and heated to 2300 K. (a) Is the system at equilibrium? If the system is not at equilibrium, in which direction must the reaction proceed to reach equilibrium? The system is at equilibrium. OThe system is not at equilibrium. The reaction must proceed to the left. The system is not at equilibrium. The reaction must proceed to the right (b) Calculate the equilibrium concentrations of all three substances. (Enter unrounded answers.) [02 - [NO M

Suppose a 250. mL flask is filled with 1.7 mol of NO2, 0.80 mol of CO and 0.70 mol of CO2. The following reaction becomes possible: NO2(g) + CO(g) = NO(g) + CO2(g) The equilibrium constant K for this reaction is 2.90 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

What is the correct classification of the following reaction? H2C=CH2 B) Oxidation reaction O C) Elimination reaction O D) Substitution reaction A) Reduction reaction

O KINETICS AND EQUILIBRIUM Calculating equilibrium composition from an equilibrium constant Suppose a 250. mL flask is filled with 1.8 mol of NO3 and 1.1 mol of NO2. The following reaction becomes possible: NO₂(g) + NO(g) + 2NO₂(g) The equilibrium constant K for this reaction is 0.707 at the temperature of the flask. Calculate the equilibrium molarity of NO3. Round your answer to two decimal places. M × S 1/5