Suppose a 500. mL flask is filled with 0.40 mol of N, and 0.90 mol of NH3. This reaction becomes possible: N,(g) + 3H,(g) → 2 NH, (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 N,. You can leave out the M symbol for molarity. N2 H, NH, initial change equilibrium Dlo

Suppose a 500. mL flask is filled with 1.5 mol of SO, and 0.20 mol of SO2. This reaction becomes possible: 2 (3) 2S0, (g) + 0,(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 0,. You can leave out the M symbol for molarity. 2 so, 02 so, initial ? change equilibrium olo

Suppose a 250. mL flask is filled with 1.5 mol of CO, 0.80 mol of NO and 0.70 mol of CO,. The following reaction becomes possible: 2' NO, (g) +CO(g) - NO(g)+CO,(g) The equilibrium constant K for this reaction is 6.92 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. olo Ar OM ?

"Synthesis gas" is a mixture of carbon monoxide and water vapor. At high temperature synthesis gas will form carbon dioxide and hydrogen, and in fact this reaction is one of the ways hydrogen is made industrially. A chemical engineer studying this reaction fills a 25 L tank with 11. mol of carbon monoxide gas and 12. mol of water vapor. When the mixture has come to equilibrium he determines that it contains 6.9 mol of carbon monoxide gas, 7.9 mol of water vapor and 4.1 mol of hydrogen gas. The engineer then adds another 3.0 mol of water, and allows the mixture to come to equilibrium again. Calculate the moles of carbon dioxide after equilibrium is reached the second time. Round your answer to 2 significant digits. mol

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

Suppose a 500. mL flask is filled with 0.30 mol of H, and 1.4 mol of HI. This reaction becomes possible: H,(2) +I,(g) = 2HI(s) 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 I,. You can leave out the M symbol for molarity. H, HI initial change equilibrium Continue Submi O 2021 McGraw-Hill Education. All Rights Reserved. Terms of Use Priva

Suppose a 500. mL flask is filled with 0.50 mol of N, and 0.90 mol of NO. This reaction becomes possible: N,(8) +O,(g) – 2NO(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 N,. You can leave out the Msymbol for molarity. NO initial change equilibrium

Suppose a 250. mL flask is filled with 1.1 mol of NO3 and 2.0 mol of NO. The following reaction becomes possible: NO3(g) + NO(g) → 2NO,(g) The equilibrium constant K for this reaction is 8.88 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

Suppose a 250. mL flask is filled with 1.6 mol of I, and 1.3 mol of HI. This reaction becomes possible: H,(g) +I,(g) = -2HI(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 H,. You can leave out the M symbol for molarity. H, I, HI initial change equilibrium olo X

Write the equilibrium 6H2O(g) + 6CO2(g) C6H12O6(s) + 6O2(g)