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

Question

The picture below shows two bulbs connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.900 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 degree C. 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? NO O_2 NO_2 What are the partial pressures of the gases? If the gas was consumed completely, put 0 for the answer PO_NO = P_O_2 = P_NO_2 =

Explanation / Answer

first find number of moles of each gas

use ideal gas equation to find moles of gas

PV = nRT

{P = pressure, V = volume, n = no. of moles, R = gas constant (0.082 L atm/mol K), T = temperature (22+273 K=295 k) }

moles of NO = PV / RT

moles = (0.9 atm x 6.0 L / 0.082 x 295 K) = 0.2232 moles

moles of Oxygen = (2.50 atm x 1.5 L / 0.082 x 295 K) = 0.155 moles

balence chemical reaction:-

2NO (g) + O2 (g) -----> 2NO2 (g)

from balence reaction 2 mole of NO react with 1 mole of O2 and oproduce 2 mole of NO2

so, 0.2232 moles of NO need = (0.2232 / 2 ) = 0.1116 moles of O2

so , O2 is in excess amount after reaction complet

O2 = (0.155 - 0.1116) moles = 0.0434 moles

O2 and NO2 will be pressent

after end of the experiment NO are not pressent so partial pressure due to NO will be = 0 atm

partial pressure due to O2 and NO2

partial pressure (P1) = n1RT/V

V = 6.0 L + 1.5 L = 7.5 L

PO2 = (0.0434 x 0.082 x 295 K) / 7.5 L = 0.13998 ~ 0.14 atm

PNO2 = (0.2232 x 0.082 x 295 k ) / 7.5 L = 0.71989 atm ~ 0.72 atm

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