Carbon dioxide and water react to form methanol and oxygen, like this: CO2(g) + H2O (g) > CH3OH (l) + O2 (g) At a certain temperature, a chemist finds that 8.6 L a reaction vessel containing a mixture of carbon dioxide, water, methanol, and oxygen at equilibrium has the following composition: compound amount CO2 2.25 gH2O 2.72 gCH3OH 3.82 gO2 1.98 gCalculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

68

Explanation:

First, we will calculate the molar concentration (M) of each substance using the following expression.

M = mass of the substance / molar mass of the substance × volume of solution

CO₂

M = 2.25 g / 44.01 g/mol × 8.6 L = 0.0059 M

H₂O

M = 2.72 g / 18.02 g/mol × 8.6 L = 0.018 M

CH₃OH

M = 3.82 g / 32.04 g/mol × 8.6 L = 0.014 M

O₂

M = 1.98 g / 32.00 g/mol × 8.6 L = 0.0072 M

Let's consider the following reaction at equilibrium.

CO₂(g) + H₂O(g) ⇄ CH₃OH(l) + O₂(g)

The concentration equilibrium constant (Kc) will not include CH₃OH because it is a pure liquid. Then,

Kc = [O₂] / [CO₂] × [H₂O]

Kc = 0.0072 / 0.0059 × 0.018

Kc = 68

answer: they produce more slow than unicellular organisms.