Nitric oxide (NO) reacts with molecular oxygen as follows:
2NO(g) + O2(g) �¨ 2NO2(g)
Initially NO and O2 are separated as shown below. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end of the reaction and calculate their partial pressures. Assume that the temperatures remains constant at 25 Celsius.
Calculate the partial pressure of NO, O2 and NO2.
Volume of NO= 4.00L
Pressure of No= 0.500atm
Pressure of O2= 1.97 atm
Volume of O2 = 1.00 L
if equilibrium partial pressure of n2, o2 and no are respectively at 2200 degrees C, what is kp?
To determine the gases that remain at the end of the reaction and calculate their partial pressures, we need to apply stoichiometry and ideal gas law equations.
First, let's calculate the number of moles of each gas present initially:
For NO:
Using the ideal gas law equation: PV = nRT
n(NO) = (PV)/(RT) = (0.500 atm * 4.00 L)/(0.0821 L·atm/mol·K * 298 K) = 0.080 mol
For O2:
n(O2) = (PV)/(RT) = (1.97 atm * 1.00 L)/(0.0821 L·atm/mol·K * 298 K) = 0.082 mol
Since the stoichiometric ratio between NO and O2 is 2:1, the reaction will consume all the O2 and produce twice the amount of NO2.
Now, let's see how much NO2 is produced and how much NO is remaining:
1. The reaction consumes 1 mole of O2, so it will produce 2 moles of NO2.
2. As stated initially, we have 0.080 mol of NO. Since the stoichiometric ratio is 2:2 (2 moles of NO will react to produce 2 moles of NO2), we can say that 0.080 moles of NO will produce 0.080 moles of NO2.
So, the total amount of NO2 produced is 2 * 0.080 = 0.160 moles.
Since we stated that the volume remained constant throughout the reaction:
The partial pressure of NO2 is calculated by using the ideal gas law:
P(NO2) = (n(NO2) * RT) / V = (0.160 mol * 0.0821 L·atm/mol·K * 298 K) / 4.00 L = 0.387 atm
Since NO2 is completely consumed, it is not present at the end.
The partial pressures of NO and O2 at the end are equal to the remaining amounts of each gas multiplied by the ideal gas constant:
P(NO) = (n(NO) * RT) / V = (0.080 mol * 0.0821 L·atm/mol·K * 298 K) / 4.00 L = 0.195 atm
Since NO was only partially consumed and not in a stoichiometric ratio of 2:1 with O2, it remains at a partial pressure of 0.195 atm.
P(O2) = (n(O2) * RT) / V = (0.082 mol * 0.0821 L·atm/mol·K * 298 K) / 1.00 L = 1.95 atm
Since O2 was completely consumed, it is not present at the end.
Therefore, at the end of the reaction, the remaining gases and their partial pressures are:
NO with a partial pressure of 0.195 atm
O2 is completely consumed and not present
NO2 is completely consumed and not present