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 6.00L contains nitric oxide at a pressure of 0.700 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 °C. After the stopcock is opened, the gases mix and react No 2NO(g) +02(g) 2NO2(g) Which gases are present at the end of the experiment? NO 2 What are the partial pressures of the gases? If the gas was consumed completely, put 0 for the answer. Number Number Number atm atm atm NO NO

Explanation / Answer

Moles of NO = [(0.700 atm) x (6 L)]/ [(0.0821 L-atm/ mol-K) x (295.15 K)]
= 0.173 moles
Moles of O2 = [(2.50 atm) x (1.5 L)]/ [(0.0821 L-atm/ mol-K) x (295.15 K)]
= 0.155 moles
Now as per the given reaction: 2NO + O2 ----> 2NO2
2 mole of NO reacts with 1 mole O2 to give 2 moles of NO2.
Therefore, 1 mole of NO requires 0.5 mole of O2.
An finally 0.173 mole NO requires = 0.5 x 0.173 = 0.0865 moles O2.
However moles of O2 available are 0.155 moles. Therefore, O2 is in excess and NO is the limiting reagent. Therefore if the reaction went to completion the NO will be exhausted completely and the gases available at the end of the experiment will be O2 and NO2.
Since NO is completely consumed the partial pressure of NO, PNO will be 0 atm.
Now the pressure due to O2, PO2 will be = (0.155- 0.0865 mole) x (0.0821 L-atm/ mol-K)x (295.15 K)]/ ( 6.0 + 1.5 L)
= 0.2213 atm.
Now the pressure due to NO2, PNO2 will be = (0.173 mole) x (0.0821 L-atm/ mol-K)x (295.15 K)]/ ( 6.0 + 1.5 L)
= 0.5589 atm.

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