Question

. A 10.0-L container is filled with 0.10 mol of H2(g) and heated to 3050 K. At that temperature, some of the H2(g) decomposes into H(g), and the total pressure is found to be 3.0 atm. Find the partial pressure of the H(g) that has been formed. (Keep in mind that there is a reaction taking place, so as H(g) is being formed, H2(g) is being consumed.)

Answer 1

Answer:

The partial pressure of the H(g) that has been formed is 1 atm.

Explanation:

Hi, we know that originally we had 0.10 mol of H2(g), but as the reaction progresses, the number of mol in the container changes. So, first, we need to check what is the number of mol in the container when pressure is 3 atm. That is:

$n=\frac{PV}{RT}$

Where:

P=3 atm

V=10L

R= constant

T=3,050K

This should look like this:

$n=\frac{(3atm)(10L)}{(0.082\frac{atm.L}{K.mol} )3,050K}=0.12mol$

We also know that H2(g) decomposes as follows:

(Equation 1.)

$H_{2} -->2H$

we also know that inside the container there are only mol of H2 and H, therefore:

(Equation 2.)

$H_{2} +H=0.12$

Substitute H2 for 2H and we get

$2H+H=0.12\\3H=0.12\\H=\frac{0.12}{3} \\H=0.04mol$

Now, the partial preasure for H would be equals to

$P(H)=\frac{n(H)}{n(total)} P_{total}$

$P(H)=\frac{0.04}{0.12)} 3atm=1atm$

So, the partial preasure of H(g) is 1 atm.

Best of luck.