This is a repost, but I have a question.
2.) For the reaction below, initially 3.5 mol of NH3 are placed in a 4.0 L
reaction chamber. After 3.0 minutes only 1.6 moles of NH3 remain.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
a. Calculate the average rate of reaction with respect to NH3
b. Calculate the average rate at which H2O is being formed.
c. Calculate the average rate at which O2 is being consumed.
"the reaction rate for NH3 is 1/4*0.158 as the defined rate of reaction for NH3. Remember the negative sign for reactants and positive sign for formation of products."
I understand what you mean by "Remember the negative sign for reactants and positive sign for formation of products." I'm confused to how I can get another negative to cancel each other and become positive in the products can you give an example and explain please)?
Also, Why did you divide NH3 by 4? "NH3 is 1/4*0.158 as the defined rate of reaction for NH3." Can you elaborate?
Also if reactants are negative wouldn't that mean that my final answer for my average rate of both NH3 and O2 would be negative as well? example: -1(1/4*0.158 )?
First question about + and - signs. I think most of the answer is "by definition" but the explanation goes like this.
In the equation A + B ==> C + D, the rate at which the reaction proceeds can be described in terms of the disappearance of A and B as rate = -d[A]/dT or -d[B]/dT or the rate of appearance of the products as d[C]/dT or d[D]/dT. Since we are talking about the rate of the FORWARD reaction you KNOW that A and B are being consumed and C and D are being formed; therefore, you want the rate of the forward reaction to be positive. That - sign in front of d[A]/dT (making it -d[A]/dT) makes that a positive number since the [A] is DESCREASING. Personally I think all of this hyperbole is nonsense and it hinges, in my opinion, on starting by talking about disappearance of reactants and appearance of products. But then I didn't make the rules. What I do is to subtract the small number from the large number (which I know will be positive) and assign the proper sign to it depending upon the subject; i.e., it is appearanve, disappearance, etc. Gotta be careful as a student, though, because it CAN lose points by a strict teacher. Hope this helps.
The next question is harder but here goes. For the reaction of
4NH3 + 5O2 ==> 4NO + 6H2O
Let me post this and I'll get to it a little later. You can think about the above answer while I eat lunch.Thanks.
ok so, -1/4 [ΔNH3] / ΔT = positive, since - ( - 0.00264) x 1/4 ? (I turn mins to seconds and divided the M) .... ? Thanks for your first explanation.
OK. Next question about dividing the NH3 by 4. Excuse me for I may go overboard but stay with me. I will use SEVERAL ways of doing this in the hopes that one will work.
4NH3 + 5O2 ==> 4NO + 6H2O
The easy way is to remind everyone that we want a set of numbers that will give us the rate of the reaction no matter which substance we choose to measure; i.e. we want to be able to express the rate using NH3 or O2 or NO or H2O and have those numbers to be consistent. So basically we are dividing the NH3 by the coefficient of 4 to make that on a per mole basis. If we were measuring the change in O2 we would divide by 5 to make that per mole. Etc. I find an example helps. If you're from the U.S. you know that two 5 cent pieces (nickels) makes a dime (a 10 cent piece). If you aren't from the U. S. think of two coins that have a 2:1 ratio value.
Suppose we have two people spending money and we want to express the rate at which each spends money but we want to have the same standard. So person 1 (p1) has a pocket full of nickels and person 2 (p2) has a pocket full of dimes. We find that p2 is spending dimes at the rate = 1 dime every minute so rate p2 = 1 dime/min. Then we find p1 is spending nickels at the rate of 2 nickels/min. We want to compare the cost of the spending (the reaction in the chemical sense) on the same basis.
The equation would look like this: 2p1 + p2 = total cost
(2*5) + (1*10) = 20 cents
So if we find p2 is 1 10cent/min our rate is 1/min for dimes. Our rate for p1 is 2 nickels/min SO WE DIVIDE THE NICKELS BY 2 (that's the coefficient)TO MAKE THAT PERSON'S RATE the same in comparison on a per coin basis. If we don't do that then the p1 spending rate is 2*5 cents/min and p2 spending is 1*10 cents/min and that's not consistent on the coin basis. Yes, it's the same total cost/min but not the same on a coin basis.The total cost for this "reaction" is 10 cents/min for p1 and 10 cents/min for p2 or 20 cents/min for the two of them but p1 has spent twice the number of coins. If we want change in the number of coins to be the same we must divide the nickels by 2 (the coefficient in the chemical sense) to make that happen. So in the case of the NH3 reaction we divide NH3 by 4 to put that on a per mole basis, we divide -d[O2]/dt by 5 or d[H2O]/dt by 6 etc. Think about it this way. If [NH3] is decreasing by 0.158 M/min and we divide by 4 to get the rate of 0.158/4 = 0.0395 M/min then we know that O2 is decreasing by 5*0.0395 M/min and that H2O is increasing by 6*0.0395 M/min.
I think that what makes this concept so difficult is that most texts use almost the same term for d(A)/dt and "decrease in (A)". I see that 0.158 called the rate of decrease and they don't distinguish very well between rate of decrease (0.158) and rate of reaction (0.158/4). In fact I see those terms used interchangeably. It confuses me too. Finally, it may be the simplest and the most straight forward way. Consider the reaction of
2H2 + O2 --> 2H2O. So we know the rate of the reaction, a made up number, is
rate O2 = -d[O2]/dt = 0.02 M/s. So we know that if 0.02 mols O2 are consumed/sec (which sounds like an easy way to express moles of consumption), then we know that hydrogen is being consumed at twice that rate. That rate will be 0.04 moles/s. If we are asked for the rate of the 2H2 + O2 ==> 2H2O reaction and you answer 0.04 moles/s and I answer 0.02 moles/s and the teacher gets us in front of the class and asks to explain. You confidently state that your number is based on 2 mol of H2 and I say my answer is based on 1 mol O2. Both of us are right; however, the teacher explains that we don't want to go through life having to explain how many moles we're basing out answer on. It makes little sense to multiply O2 by 2 so the reaction is based on 2 mols H2 OR 2 mols O2. In this case it makes more sense to base it on the smallest coefficient of which is 1 for O2 SO to be consistent we always base the question on a 1 mole basis. When you and I do the same problem, then the 0.02 is the standard, 0.04/2mols H2 = 0.02 for 1 mol H2 and it makes it unnecessary to specify henceforth how many moles our answer is based on. IT JUST TURNS OUT, AND YOU MIGHT KNOW IT WOULD, that this reaction of 4NH3 + 5O2 ==> 4NO + 6H2O HAS NO COEFFICIENT OF 1 SO IT makes it so difficult to explain in terms of using a mole. Therefore, in my first examples above I used the expression of "per mole". Here is the clincher. BY DEFINITION, the rate is
rate = -(1/4)*d[NH3]/dt = -(1/5)*d[O2]/dt = (1/4)*d[NO]/dt = (1/6)*d[H2O]/dt
That incorporates the - sign, the coefficients, the rate of consumption and the rate of formation into one neat package. Hope this makes sense to you. I have struggled to put all of this into words and keep it consistent. Isn't chemistry exciting? and fun?
To determine the average rate of reaction with respect to NH3, we can use the change in moles of NH3 over a specific time interval. In this case, we are given that 3.5 moles of NH3 initially react, and after 3.0 minutes, only 1.6 moles of NH3 remain.
The average rate of reaction with respect to NH3 can be calculated using the formula:
Average Rate = (Change in Concentration of NH3) / (Change in Time)
So, we have:
Change in Concentration of NH3 = 3.5 moles - 1.6 moles = 1.9 moles
Change in Time = 3.0 minutes
Average Rate of Reaction with respect to NH3 = (1.9 moles) / (3.0 minutes)
To explain why we divide NH3 by 4, it is based on the stoichiometry of the balanced chemical equation. The stoichiometric coefficients in the balanced equation tell us the molar ratio of reactants and products involved in the reaction.
In this case, the balanced equation is:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
From the equation, we can see that the coefficient of NH3 is 4. This means that for every 4 moles of NH3 that react, we will produce 4 moles of NO.
To compare the average rate of reaction for NH3 and NO, we need to consider the stoichiometric ratio between them. Since the coefficient of NH3 is 4, the ratio of the rate of reaction of NH3 to NO is 1:4.
Hence, we divide the rate of NH3 by 4, as stated in the provided answer, to obtain the rate of reaction for NH3 compared to the defined rate of reaction for NH3.