The reaction AB(aq)->A(g)+ B(g)is second order in AB and has a rate constant of 0.0282 M^{-1} s^{-1} at 25 Celsius. A reaction vessel initially contains 250 mL of 0.105 M AB which is allowed to react to form the gaseous product. The product is collected over water at 25 Celsius.

1-How much time is required to produce 115.0 mL of the products at a barometric pressure of 779.9 mmHg. (The vapor pressure of water at this temperature is 23.8 mmHg.)

is the answer 1392 s?

Well, it seems like you need some calculations. But before we dive into the world of numbers, let me tell you something a little more lighthearted!

Why did the math book look so sad?

Because it had too many problems to solve and not enough time for fun!

Okay, now let's get down to business. We first need to consider the effect of vapor pressure on the reaction. Since the product is being collected over water, the vapor pressure of water needs to be taken into account.

Using Dalton's Law, we subtract the vapor pressure of water from the total pressure to get the partial pressure of the product:

Partial pressure of product = Total pressure - Vapor pressure of water
= 779.9 mmHg - 23.8 mmHg
= 756.1 mmHg

Now, we can use the ideal gas law to calculate the number of moles of the product produced:

PV = nRT

n = PV / RT

Where:
P = Partial pressure of product = 756.1 mmHg * (1 atm / 760 mmHg) = 0.995 atm
V = Volume of product = 115.0 mL = 0.115 L
R = Ideal gas constant = 0.0821 L atm / (mol K)
T = Temperature = 25 Celsius = 298 K

Plugging these values in, we get:

n = (0.995 atm * 0.115 L) / (0.0821 L atm / (mol K) * 298 K)
= 0.01577 mol

Now that we know the number of moles of the product, we can determine the time required to produce this amount.

The reaction is second-order in AB, so we can use the integrated rate law for a second-order reaction:

1 / [AB]t - 1 / [AB]0 = kt

Where:
[AB]t = Concentration of AB at time t
[AB]0 = Initial concentration of AB
k = Rate constant

Rearranging the equation to solve for time (t):

t = 1 / (k * ([AB]0 - [AB]t))

Now, let's plug in the values:

k = 0.0282 M^(-1) s^(-1)
[AB]0 = 0.105 M
[AB]t = [AB]0 - (0.01577 mol / 0.250 L)
= 0.105 M - 0.06308 M
= 0.04192 M

t = 1 / (0.0282 M^(-1) s^(-1) * 0.04192 M)
= 884.47 seconds

So, it would take approximately 884.47 seconds to produce 115.0 mL of the product at a barometric pressure of 779.9 mmHg.

To answer this question, we can use the integrated rate equation for a second-order reaction, which is:

1/[AB]t - 1/[AB]0 = kt

Where:
[AB]t is the concentration of AB at time t
[AB]0 is the initial concentration of AB
k is the rate constant
t is the time

First, let's calculate the initial concentration of AB:

[AB]0 = 0.105 M

Now we need to convert the initial volume of AB to liters:

V0 = 250 mL = 0.250 L

Next, let's calculate the final concentration of AB:

[AB]t = (0.105 M * 0.250 L) / (0.250 L + 0.115 L)

To determine the concentration of AB when 115.0 mL of product has been produced, we need to consider the volume change due to the gaseous product. For a reaction at constant temperature and pressure, the volume of gaseous reactants and products is directly proportional to their molar amounts. Since the reaction is 1:1, the volume of AB will decrease by 115.0 mL.

Substituting the values into the integrated rate equation:

1/[AB]t - 1/[AB]0 = kt

1/[AB]t - 1/(0.105 M) = (0.0282 M^-1 s^-1) * t

To solve for t, we rearrange the equation:

1/[AB]t = (0.0282 M^-1 s^-1) * t + 1/(0.105 M)

Now we can plug in the values:

1/(0.105 M * (0.250 L - 0.115 L)) = (0.0282 M^-1 s^-1) * t + 1/(0.105 M)

Simplifying the equation:

1/(0.105 M * 0.135 L) = (0.0282 M^-1 s^-1) * t + 1/(0.105 M)

1/(0.014175 M L) = (0.0282 M^-1 s^-1) * t + 9.5238 M^-1

69.9657 s/M = (0.0282 M^-1 s^-1) * t

Now we can solve for t:

t = 69.9657 s/M / 0.0282 M^-1 s^-1

t = 2483.51 seconds

Finally, we convert the time to minutes:

t = 2483.51 seconds * (1 minute / 60 seconds)

t ≈ 41.39 minutes

Therefore, it would take approximately 41.39 minutes to produce 115.0 mL of the product at the given conditions.

To determine how much time is required to produce 115.0 mL of the products, we can use the integrated rate equation for a second-order reaction:

1/([AB]t) - 1/([AB]0) = kt

Where [AB]t is the concentration of AB at time t, [AB]0 is the initial concentration of AB, k is the rate constant, and t is the time.

1. Convert the initial volume of AB (250 mL) to initial concentration:
[AB]0 = 0.105 M

2. Convert the product volume from mL to L:
[AB]t = (115.0 mL / 1000) L = 0.115 L

3. Convert the barometric pressure from mmHg to atm:
Barometric pressure = 779.9 mmHg - 23.8 mmHg = 756.1 mmHg
Barometric pressure in atm = 756.1 mmHg / 760 mmHg/atm

4. Calculate the concentration of AB at time t:
1/0.115 - 1/0.105 = (0.0282 M^-1 s^-1) * t

5. Solve for t:
0.115 / 0.105 - 1 = (0.0282 M^-1 s^-1) * t

Simplifying:
1.095 - 1 = (0.0282 M^-1 s^-1) * t
0.095 = (0.0282 M^-1 s^-1) * t

Rearranging:
t = 0.095 / 0.0282 s^-1
t ≈ 3.369 s

Therefore, it would take approximately 3.369 seconds to produce 115.0 mL of the products at a barometric pressure of 779.9 mmHg.

I would do this.

Use P = MRT
where P: = (779.9-23.8)/760
and solve for M which should give A and B concns (I assume 0.115L is the volume of the gases).
Take 1/2 that to find M needed for AB and apply the second order equation of
1/A - !/Ao = akt
You know Ao, a = 1, you know k and A (from the calculation). Solve for t which should be i s since k is in s.