Consider the reaction:
2 HBr (g) → H₂ (g) + Br₂ (g)
c. If the volume of the reaction vessel in part b was 1.25 L, what amount of Br₂ (in moles) was formed during the first 15.0 s of the reaction?

Question

Consider the reaction:

2 HBr (g) → H₂ (g) + Br₂ (g)

c. If the volume of the reaction vessel in part b was 1.25 L, what amount of Br₂ (in moles) was formed during the first 15.0 s of the reaction?

Accepted Answer

Hey everyone in this example, we need to consider the given reaction, which shows the synthesis of gaseous water in a two liter flask. We're told that the concentration of our hydrogen gas went from .700 moler 2.330 moller. During the 1st 40 seconds of the reaction. We need to calculate the mass of gaseous water produced in the 1st 20 seconds of our reaction. So our first step is to write out a rate expression where we're going to take the change in concentration of our hydrogen gas according to change in time and according to our given equation, we have a coefficient of two. However, we want this to be in the denominator and because this is a reactant and we're losing this, we're going to be Making this a negative coefficient of 1/2. We're then going to set this equal to our change in concentration of our second reactant oxygen gas divided by the change in time. Where we have a coefficient of one according to our given equation. So we would have negative one times this expression here and then. Now we're going to set this equal to on our product side, where we would have the change in concentration of our water divided by change in time. Where according to our given equation, we have a coefficient of two in front of water. So we would have to being in the denominator and we're going to leave this coefficient as positive one half due to the fact that this is a product and it's being produced. So now the next step is to go ahead and take our rate And find the value for this rate based on the info they give us in the prompt for H2. So according to the prompt we have well according to our rate expression actually we have that coefficient of one half. And in our numerator we have the change in concentration of H two which according to the prompt went from 20.700 moller 2.300 moller we want to do final minus initial, so 0.300 moller is our final value and 0.700 moller is our initial value. So we would have a 0.330 moller minus 0.700 moller for final minus initial. And then in our denominator we want the change in time And so that would be 40 seconds zero seconds for our final time minus our initial time. And so what we should have here is a value for rate equal to 4.63 times 10 to the negative third power. And we're left with units of polarity per second. Now our next step is to call that polarity can be interpreted as moles divided by leaders and we want to find now the change in concentration of our product water and we will do so by taking our rate value that we just calculated multiplying it by the change in time and then dividing this by one half because we have that coefficient of one half as we stated above in front of our water since it's a product. And so what we should have when we figure out the value here Is that the change in concentration of our water is equal to our rate, which we found is 4.63 times 10 to the negative third power polarity per second. And we can actually interpret these these units as now moles divided by leaders. Time seconds. Where we're now going to multiply by the difference in time. So we should have the final time being 20 seconds for water according to the prompt minus the initial time at zero. And then this is also zero seconds. So then we're going to divide this by our coefficient in front of water from our balanced equation, which was to and we want that in the denominator. So we said that that is positive one half since water is a product. So to continue solving, we're going to simplify this and what we're going to get is a value of zero 0.1852. And as far as units will be able to cancel out our units of seconds as well as what we're left with just units of moles per liter actually. So we should have units of moles per leader and we want to recall that from the prompt, this reaction was happening in a two liter flat. So we're going to multiply this by two leaders to also cancel out our units of leaders leaving us with moles. And so what we're going to get when we multiply this by two leaders is a value of .3704 moles of our water, which now we want to go ahead and utilize the molar mass Of water from our periodic tables, which we would see as a value of 18.02 grams per mole. We're going to use this as a conversion factor to cancel out moles. So again, we said, we have 18.02g of water For one mole of water. And so now we're able to cancel out moles here and now we can can or we can calculate our massive water, which is what the prompt wants, Which will give us a value of 6.67 g of water. So this is our massive water produced in the 1st 20 seconds of our reaction. And this here is going to be our final answer to complete this example. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Consider the reaction:
2 HBr (g) → H₂ (g) + Br₂ (g)
c. If the volume of the reaction vessel in part b was 1.25 L, what amount of Br₂ (in moles) was formed during the first 15.0 s of the reaction?

Verified Solution

Key Concepts

Stoichiometry

Ideal Gas Law

Reaction Rate

Consider the reaction:2 HBr (g) → H2 (g) + Br2 (g)c. If the volume of the reaction vessel in part b was 1.25 L, what amount of Br2 (in moles) was formed during the first 15.0 s of the reaction?

Verified Solution

Key Concepts

Stoichiometry

Ideal Gas Law

Reaction Rate

Consider the reaction:
2 HBr (g) → H₂ (g) + Br₂ (g)
c. If the volume of the reaction vessel in part b was 1.25 L, what amount of Br₂ (in moles) was formed during the first 15.0 s of the reaction?