Question

co mWhen CO2( g) is put in a sealed container at 701 K and a pressure of 10.0 atm and is heated to 1401 K, the pressure rises to 22.5 atm. Some of the CO2 decomposes to CO and O2. Calculate the mole percent of CO2 that decomposes.

Answer 1

25.06%

Step-by-step explanation:

For the ideal gas law, we can calculate the number of moles of the gas:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature.

The volume must be constant because it's a sealed container. So, in the beginning:

10*V = n*0.082*701

57.482n = 10V

n = 0.1740V

After de decomposition:

22.5*V = n'*0.082*1401

114.882n' = 22.5V

n' = 0.1958V (which is the total amount of moles of the gases)

Let's supose, V = 1L, so:

n = 0.1740 mol, and n' = 0.1958

For the decomposition reaction, we can do a reaction table:

2CO₂(g) → 2CO(g) + O₂(g)

0.1740 0 0 Initial

-2x +2x +x Reacts( stoichiometry is 2:2:1)

0.1740-2x 2x x Final

n' = 0.1740 - 2x + 2x +x

0.1740 + x = 0.1958

x = 0.0218 mol

So, 0.0436 (2*0.0218) mol of CO₂ decomposes, the percent is:

(0.0436/0.1740)*100% = 25.06%