What is the pressure of the C2H5OH vapor at 650k before it decomposes?

Question

92.5kj + C2H5OH<->H2+CO +CH4

A 15.50-gram sample of C2H5OH liquid is placed into a 4.50-liter flask and heated to 650.0 k. The sample is vaporized, and then begins to decompose. What is the pressure of the C2H5OH vapor at 650k before it decomposes?

Answer 1

At first glance, this problem seems a lot harder than it actually is. You are given in the problem the following:

V = 4.5 L

T = 650 K

m = 15.5 g C₂H₅OH.

This is enough to solve the problem.

Step 1. Find the molar mass of C₂H₅OH.

C 2 x 12 = 24 g/mol

H 5 + 1 = 6 x 1 = 6 g/mol

O. 1 x 16 = 16 g/mol

Total: 46 g/mol.

Next convert 15.5 g C₂H₅OH into mol C₂H₅OH by dividing 15.5 g / (46 g/mol) = 0.33 mol

Step 2. Use the ideal gas law formula to solve for pressure. PV = nRT ==> P = nRT/V

Plug in.

P = (0.33 mol) * (0.08205 Latm/molK) * (650 K) / (4.5 L) = 3.9 atm

The answer is P = 3.9 atm.

At this temperature, the equilibrium will favor overcoming the activation energy (Enthalpy in the reaction), so the forward reaction will almost spontaneously go to products and the decomposition of the C₂H₅OH will occur. Just prior to decomposition of the C₂H₅OH, only C₂H₅OH will be present in the container (due to the vaporization), so all of the pressure will be due to the C₂H₅OH in vapor (gas) form.