Nitrogen and oxygen react to form nitric oxide (NO). All substances are in the gas phase. If 0.352 atm of nitrogen and 0.511 atm of oxygen react, what is the partial pressure of nitric oxide (in mmHg) when this reaction goes 73.9 complete. The temperature and volume are constant.
To find the partial pressure of nitric oxide (NO) when the reaction is 73.9% complete, we need to use the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction:
N₂ + O₂ → 2NO
According to the stoichiometry of the reaction, one mole of N₂ reacts with one mole of O₂ to produce two moles of NO.
To find the moles of NO produced, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product formed.
To determine the limiting reactant, we compare the ratios of the reactants to the balanced equation coefficients.
For nitrogen (N₂):
0.352 atm × (1 mol N₂/1 atm) = 0.352 mol N₂
For oxygen (O₂):
0.511 atm × (1 mol O₂/1 atm) = 0.511 mol O₂
Since the balanced equation has a ratio of 1:1 for N₂ and O₂, we see that the limiting reactant is nitrogen (N₂) because there are fewer moles of nitrogen present.
Now, we can determine the moles of nitric oxide (NO) produced using stoichiometry. Since one mole of nitrogen reacts to form two moles of nitric oxide, we have:
0.352 mol N₂ × (2 mol NO/1 mol N₂) = 0.704 mol NO
The reaction is 73.9% complete, which means that 73.9% of the limiting reactant has been consumed. Therefore, the moles of NO produced at this point would be:
0.704 mol NO × 0.739 = 0.519 mol NO (rounded to three significant figures)
Finally, we can calculate the partial pressure of nitric oxide (NO) in mmHg using the ideal gas law. The ideal gas law equation is:
PV = nRT
where:
P is the pressure,
V is the volume (which is constant in this case),
n is the number of moles,
R is the ideal gas constant (0.0821 L·atm/(mol·K)), and
T is the temperature (which is constant in this case).
In this equation, we can solve for P by rearranging it as:
P = (n * R * T) / V
Since the volume and temperature are constant, we can simplify the equation to:
P = (n * R) / V
Now, we can substitute the values into the equation:
P(NO) = (0.519 mol * 0.0821 L·atm/(mol·K)) / V
However, since we are looking for the partial pressure of nitric oxide (NO) in mmHg, we should convert the pressure unit from atm to mmHg using the conversion factor:
1 atm = 760 mmHg
Therefore, the equation becomes:
P(NO) = (0.519 mol * 0.0821 L·atm/(mol·K)) / V * (760 mmHg/1 atm)
To complete the calculation, we need to know the value of the volume (V). If you provide the volume, we can substitute it into the equation to find the partial pressure of nitric oxide (NO) in mmHg.
I assume you mean 73.9% complete.
N2 + O2 ==> 2NO
This is a limiting reagent (LR) problem. First we determine the LR.
atm NO formed if we used all of the N2 and an unlimited supply of O2.
0.352 atm N2 x (2 mols NO/1 mol N2) = 0.352 x 2/1 = 0.704 atm NO produced.
atm NO formed if we used all of the O2 and an unlimited supply of N2.
0.511 atm O2 x (2 mol NO/1 mol O2) = 1.022 atm NO produced.
You can see these two answers don't agree so one of them must be wrong. In LR problem, the correct value is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Therefore, N2 is the limiting reagent and we will produce 0.704 atm NO at 100% yield. Since this is only 73.9% yield the product will be 0.704 x 0.739 = ?.