Suppose a 500. mL flask is filled with 0.90 mol of CO, 1.8 mol of CO2 and 1.4 mol of H2. The following reaction becomes possible: CO(g) + H2O(g) CO2(g) + H2(g) = The equilibrium constant K for this reaction is 0.274 at the temperature of the flask. Calculate the equilibrium molarity of H2O. Round your answer to two decimal places. Ом

Consider the equilibrium system described by the chemical reaction below. A mixture of gas containing only N₂ and H₂ is reacted in a vessel at high temperature. At equilibrium, the 5.0 M H₂, 8.0 M N₂, and 4.0 M NH 3 are present. Determine the initial concentrations of H₂ and N₂ that were present in the vessel. N₂(g) + 3 H₂(g) = 2 NH3(g) Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. Initial (M) Change (M) Equilibrium (M) -2.0 0 2.0 5.0 -5.0 N₂(g) 8.0 1 6.0 + 4.0 -6.0 3 H₂(g) -4.0 10.0 2 NH3(g) RESET 11.0

Suppose a 500. mL flask is filled with 1.8 mol of Cl,, 1.6 mol of CHCI, and 1.7 mol of HCl. The following reaction becomes possible: Cl,(g) +CHCI, (g) HCl(g)+CCl,(g) The equilibrium constant K for this reaction is 0.642 at the temperature of the flask. Calculate the equilibrium molarity of CHCI,. Round your answer to two decimal places.

Using the general properties of equilibrium constants At a certain temperature, the equilibrium constant K for the following reaction is 924.: CO(g) + H₂O(g) → CO₂(g) + H₂(g) Use this information to complete the following table. Suppose a 17. L reaction vessel is filled with 0.75 mol of CO₂ and 0.75 mol of H₂. What can you say about the composition of the mixture in the vessel at equilibrium? What is the equilibrium constant for the following reaction? Round your answer to 3 significant digits. CO₂(g) + H₂(9) CO(g)+H₂O(g) What is the equilibrium constant for the following reaction? Round your answer to 3 significant digits. 3 CO(g) + 3H₂O(g) P 3 CO₂(g) + 3H₂(9) There will be very little CO and H₂O. There will be very little CO2 and H₂. Neither of the above is true. K = 0 K = 0 ■ x10 X ? 00. Ar 8.