Question

When CO2(g)CO2(g) is put in a sealed container at 730 KK and a pressure of 10.0 atm and is heated to 1420 KK , the pressure rises to 24.1 atm. Some of the CO2CO2 decomposes to COCO and O2O2.

Calculate the mole percent of  CO2 that decomposes.

Express your answer using two significant figures.

%CO2 =

Answer 1

Ideal Gas Law:

pV=nRT

R = 0.08206 L×atm/mol×K

$10.0 \times V = n \times 0.08206 \times 730 \\ 10.0 \times V = 59.90n \\ n = 0.1669V$

After decomposition:

$24.1 \times V = n^* \times 0.08206 \times 1420 \\ 24.1 \times V = 116.52 \times n^* \\ n^* = 0.2068V$

(which is the total amount of moles of the gases)

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

$n=0.1669 \\ n^*=0.2068$

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

2CO₂ → 2CO + O₂

0.1669____ 0___ 0

-2x______ +2x__ +x

0.1669-2x __2x__x

$n^* = 0.1669 -2x + 2x +x \\ 0.1669 + x = 0.2068 \\ x = 0.0399$

So, $2 \times 0.0399 = 0.0798$ mol of CO₂ decomposes, the percent is:

$\frac{0.0798}{0.1669} \times 100 \;\% = 47.81 \; \%$

Answer: 47.81 %