Samples of NH3 and 02 are placed in a rigid container. When the equilibrium below is established, the partial pressure of H2O is 0.784 atm. What is the partial erasure of NO when the system is at equilibrium?

4NH3 + 5O2 -----> 4NO + 6H2O

Well, that's quite the chemical equation! It seems like these molecules are having a real chemical party in that rigid container. Now, you're wondering about the partial pressure of NO when the system reaches equilibrium. Let me put on my chemistry hat and see if I can crack a funny answer for you.

Since we're dealing with gases, we know that the partial pressure of each component contributes to the total pressure. In this equation, we have 4 moles of NH3 reacting with 5 moles of O2 to produce 4 moles of NO. Now, if we assume that the initial pressures of NH3 and O2 were P1 and P2 respectively, we have:

(P1 + P2 + PNO) = 0.784 atm

Since the mole ratio is 4:4, we can conveniently assume that the pressure of NO is also P1. So, we have:

(P1 + P2 + P1) = 0.784 atm
(2P1 + P2) = 0.784 atm

Now, to crack a joke, we need to calculate the values of P1 and P2. Unfortunately, I can't do that without more information. So, until I get the numerical values, I'm afraid I can't give you a specific answer. But hey, at least I can provide you with a clown's worthy chemistry explanation, right?

To determine the partial pressure of NO (nitric oxide), we need to use the balanced equation and the given information about the partial pressure of H2O.

The balanced equation is:
4NH3 + 5O2 -> 4NO + 6H2O

From the equation, we can see that the ratio of NH3 to NO is 4:4 or 1:1. This means that the partial pressure of NO will be the same as the partial pressure of NH3 at equilibrium.

Since the partial pressure of H2O is given as 0.784 atm, we can assume that the partial pressure of NH3 is also 0.784 atm.

Therefore, the partial pressure of NO at equilibrium is 0.784 atm.

To find the partial pressure of NO, we need to determine the equilibrium constant (Kp) for the given reaction and then use it to calculate the partial pressure.

Step 1: Write the expression for the equilibrium constant (Kp) using the given balanced equation:

Kp = (pNO)^4 / (pNH3)^4 * (pH2O)^6 / (pO2)^5

Step 2: Plug in the given value for the partial pressure of H2O (0.784 atm) into the equation and rearrange it:

0.784^6 = (pNO)^4 / (pNH3)^4 * (0.784)^6 / (pO2)^5

Step 3: Since the container is rigid, the sum of the partial pressures of all the gases remains constant. Therefore, we can rewrite the equation as:

(0.784 + pNO)^6 = ((pNH3 + pNO)/4)^4 * ((0.784 + pNO)/5)^5

Step 4: Simplify the equation and solve for pNO:

0.784^6 + 6*(0.784^5)(pNO) + 15*(0.784^4)(pNO)^2 + 20*(0.784^3)(pNO)^3 + 15*(0.784^2)(pNO)^4 + 6*(0.784)(pNO)^5 + (pNO)^6 = ((pNH3 + pNO)/4)^4 * ((0.784 + pNO)/5)^5

Step 5: Use a numerical method or calculator to solve the equation for pNO.

Alternatively, you can also use an online equilibrium calculator or software to input the given values and obtain the partial pressure of NO at equilibrium.

......4NH3 + 5O2 -----> 4NO + 6H2O

I......x......y..........0.....0
C....-4z.....-5z........+4z...+6z
E.....x-4z..y-5z.........4z....6z

pH2O = 6z = 0.784 from the problem
z = ?
Then NO is 4z = ?

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