The picture belaw shows two bulbs connected by a stopcock. The large bulb, with

Question

The picture belaw shows two bulbs connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0 550 atm, and the small bulb, with a volume of 1 50 L, contains oxygen at a pressure of 2 50 atm. The temperature at the beginning and the end of the experiment is 22 degreeC. After the stopcock is opened, the gases mix and react: 2NO(g) + O_2(g) rightarrow 2NO_2(g) which gases are present at the end of the experiment? What are the partial pressures of the gases? If the gas was consumed completely. Put 0 for the answer.

Explanation / Answer

we know that

for gases

PV = nRT

so

initially for N0

0.55 x 6 = n x 0.0821 x 295

n = 0.136254

so

initial moles of NO = 0.136254

now

for 02

2.5 x 1.5 = n x 0.0821 x 295

n = 0.154834

so

initial moles of 02 = 0.154834

now

consider the reaction

2NO + 02 ---> 2 N02

we can see that

moles of 02 reacted = 0.5 x moles of N0

so

moles of 02 reacted = 0.5 x 0.136254

moles of 02 reacted = 0.068127

so

only 0.068127 moles of 02 reacted and all the NO is consumed

so

finally the gases present are 02 and N02

now

moles of N02 formed = moles of NO reacted

moles of N02 formed = 0.136254

now

final volume = 2.5 + 1.5 = 4 L

moles of 02 left = 0.154834 - 0.068127 = 0.086707

so

for 02

P x 4 = 0.086707 x 0.0821 x 295

P = 0.525 atm

so

partial pressure of 02 = 0.525 atm

now

for N02

P x 4 = 0.136254 x 0.0821 x 295

P = 0.825 atm

so

the answers are

1) gases present at the end are 02 and N02

2) finally

pN0 = 0

p02 = 0.525 atm

pN02 = 0.825 atm

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