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
Chemistry - DrBob222, Sunday, October 31, 2010 at 8:08pm
2NO + O2 ==> 2NO2
mols NO =?? Use PV = nRT
moles O2 = ?? Use PV = nRT
From moles determine the identity of the limiting reagent, use that determine how much of the other reagent remains unreacted, then use PV = nRT to determine pressure of the "other" reagent and pressure o the product. Post your work if you get stuck.
Chemistry - Sarah, Sunday, October 31, 2010 at 8:28pm
PNO2= 2.00 atm x L/(4.00 L + 1.97 L)= 0.335 atm
PO2= 0.97 atm x L/ (4.00 L + 1.97 L)= 0.162 atm
PNO=????
Show step by step how do I get PNO.
First things first.
The problem states NO is 0.5 atm and 4.00 L initially. But you haven't calculated n for NO.
The problem states O2 is 1.97 atm and 1.00 L but you have used 0.97 atm and 4.00 L. And there are other errors/typos, too. I may not be interpreting your answer correctly.
How would you calculate the problem I am confuse. Please show me.
I gave you step by step the first time.
You can substitute as well as I.
To find the partial pressure of NO (PNO), we need to determine the moles of NO and the moles of O2 initially and use these values to determine the limiting reagent in the reaction.
First, let's find the moles of NO using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
Given:
Pressure of NO = 0.500 atm
Volume of NO = 4.00 L
Using PV = nRT:
n(NO) = (P(NO) * V(NO)) / (R * T)
Since we are given the pressure in atm, we need to convert the temperature to Kelvin.
Temperature in Celsius = 25°C
Temperature in Kelvin = 25°C + 273.15 = 298.15 K
Now we can calculate the moles of NO:
n(NO) = (0.500 atm * 4.00 L) / (0.0821 L·atm/mol·K * 298.15 K) = 0.0801 mol
Next, let's find the moles of O2 using the same formula but with the given values for pressure and volume of O2:
Pressure of O2 = 1.97 atm
Volume of O2 = 1.00 L
n(O2) = (P(O2) * V(O2)) / (R * T)
n(O2) = (1.97 atm * 1.00 L) / (0.0821 L·atm/mol·K * 298.15 K) = 0.0801 mol
From the balanced equation 2NO + O2 → 2NO2, we can see that the mole ratio between NO and O2 is 2:1. However, the number of moles we calculated for NO and O2 is equal, indicating that neither NO nor O2 is in excess. This means that both NO and O2 will react completely.
Therefore, the reaction will proceed to completion, resulting in the full conversion of NO and O2 to NO2. As a result, no unreacted NO or O2 will remain at the end of the reaction.
So, the partial pressure of NO (PNO) at the end of the reaction would be zero (0 atm) since there is no NO remaining.
To summarize:
PNO = 0 atm
PO2 = 0 atm
PNO2 = Calculated previously as 0.335 atm