The problem:
Nitrogen can react with steam to form ammonia and nitrogen monoxide gas. A 20.0L sample of nitrogen at 173C and 772mmHg is made to react with an excess of steam. The products are collected at room temperature (25C) into an evacuated flask with a volume of 15.0
A. Write a balanced equation for the reaction.
B. What is the total pressure of the products in the collecting flask after the reaction is complete?
C. What is the partial pressure of each of the products in the flask?
My homework:
A. N(g) + H2O(g)--> NO(g)+ NH3(g)
I also determined that:
T=173C=446.15K & 25C=298.15
P=772mmHg=1.016atm
I believe I have A. correct but B and C are giving me a problem. I am sure I could get it after some sleep but I have been doing chemistry for 10 hours straight and could really use a little more help on this last problem.
The chemical equation is:
5N2(g) + 6H2O(g)--> 6NO(g) + 4NH3(g)
The coefficients can be used either as a mole ratio or a volume ratio.
The volumes of the products are:
20.0L of N2 x (6.00 L NO / 5 L N2) = 24.0 L NO
20.0L of N2 x (4.00 L NH3 / 5 L N2) = 16.0 L NH3
TOTAL volume of products = 24.0L + 16.0L = 40.0L
The mole fractions of each product gas are:
X(NO) = 24/40 = 0.600
X(NH3) = 16/40 = 0.400
• To find the final total pressure in the flask use:
PV = nRT
• To find the final pressures of NO and NH3, multiply each mole fraction by the previous answer.
I am letting you do the last two part of the solution on your own.
Please disregard I finally figured it out.
Sure, I can help you with that! I understand that chemistry can be exhausting, so let's try to tackle this problem with a touch of humor to lighten the mood.
B. What is the total pressure of the products in the collecting flask after the reaction is complete?
Well, after all that reaction and steamy chemistry, we're left with a flask full of products. To find the total pressure, we need to take into account the pressure from the nitrogen and the pressure from the steam. But wait, we also need to consider that the steam was in excess, meaning it was overly enthusiastic about reacting. So, the steam used up all the nitrogen, leaving only the pressure from the steam.
Therefore, the total pressure of the products in the flask after the reaction is complete is equal to the pressure from the excess steam. So, the answer is 1.016 atm (since 772 mmHg is approximately equal to 1.016 atm).
C. What is the partial pressure of each of the products in the flask?
In the clown world of chemistry, we have two jokers in the flask: nitrogen monoxide gas (NO) and ammonia (NH3). To find their partial pressures, we need to know the mole ratios from the balanced equation. So, let's take a look at our balanced equation again:
N2(g) + H2O(g) --> NO(g) + NH3(g)
From the equation, we can see that for every 1 mole of nitrogen that reacts, we get 1 mole of NO and 1 mole of NH3. This means that the mole ratios and partial pressures of NO and NH3 are equal.
Since the total pressure of the products in the flask is 1.016 atm (as we determined in answer B), the partial pressure of NO and NH3 would each be 1.016 atm.
Remember, in the world of chemistry, even clowns follow the laws of mole ratios and balanced equations! I hope this helps, and keep up the good work!
I understand that you've already made good progress on this problem. Let's continue with finding the answers to part B and C.
To determine the total pressure of the products in the collecting flask after the reaction, we can use the ideal gas law equation: PV = nRT.
In this case, we have the volume (V) of the flask, which is 15.0 L, the temperature (T) of the products, which is 25°C or 298.15 K, and we need to find the pressure (P). We also need the number of moles (n) for both ammonia (NH3) and nitrogen monoxide (NO).
For part B, we assume that the reaction goes to completion, which means all of the nitrogen reacts with steam to form ammonia and nitrogen monoxide. Therefore, the moles of NH3 and NO produced will be the same as the moles of nitrogen that reacted.
To find the moles of nitrogen, we need to use the ideal gas law again, rearranged to solve for moles (n):
n = PV / RT
Substituting the given values: P = 1.016 atm, V = 20.0 L, T = 446.15 K, and R = 0.0821 L·atm/(mol·K), we can calculate the moles of nitrogen.
Now that we know the moles of NH3 and NO, we can use the ideal gas law once again to find the total pressure of the products in the flask. Since the flask is evacuated, the pressure is solely due to the products.
Now, let's move on to part C, which asks for the partial pressure of each of the products in the flask.
To find the partial pressure of NH3 and NO, we need to know their mole fractions. The mole fraction of a gas is the ratio of the moles of that gas to the total moles of gas in the flask.
Once we have the mole fractions, we can find the partial pressures using Dalton's law of partial pressures. According to Dalton's law, the total pressure in the flask is the sum of the partial pressures of each gas.
To calculate the partial pressures, we can multiply the mole fractions by the total pressure found in part B.
I hope this explanation helps you with part B and C of the problem. If you have further questions or need more assistance, feel free to ask!