suppose that a large volume of 3% hydrogen peroxide decomposes to produce 12mL of oxygen gas in 100 s at 298 K. Estimate how much oxygen gas would be produced by and identical solution in 100s at 308 K

hi can you please help me set up this problem, then I can do the math, I just don't know what formula to use. Thanks clo

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature in Kelvin

In this case, we are given the volume and temperature, and we need to find the amount of oxygen gas produced (n) at the new temperature.

First, let's calculate the number of moles of oxygen gas produced in the given conditions (298 K):

From the equation:
2H2O2(l) -> 2H2O(l) + O2(g)

The molar ratio shows that one mole of hydrogen peroxide produces one mole of oxygen gas.

Given that 12 mL of oxygen gas was produced in 100 s, we can convert it to liters and find the number of moles by using the ideal gas law equation:

V1 = 12 mL = 0.012 L
T1 = 298 K
P1 = ?
n1 = ?

Now, let's consider the new conditions (308 K) and calculate the new number of moles of oxygen gas (n2) using the ideal gas law equation:

V2 = ?
T2 = 308 K
P2 = ?
n2 = ?

To continue, we need to make an assumption about the pressure of the gas. If the pressure remains constant, we can simplify the equation to:

(P1 * V1) / T1 = (P2 * V2) / T2

Rearranging the equation to solve for V2, we have:

V2 = (P1 * V1 * T2) / (P2 * T1)

Since the volume is constant, V1 = V2. We can substitute V1 = V2 into the equation:

V1 = V2 = (P1 * V1 * T2) / (P2 * T1)

Canceling out V1:

1 = (P1 * T2) / (P2 * T1)

Now, we can solve for P2:

P2 = (P1 * T2) / (T1)

Let's calculate the new pressure (P2) at 308K:

P1 = assumed constant pressure at 298 K

Now, using the calculated P2, we can find the new number of moles (n2) of oxygen gas using the ideal gas law equation:

n2 = (P2 * V2) / (R * T2)

From the given information, the volume is constant, so V1 = V2. We can substitute V1 = V2 into the equation:

n2 = (P2 * V1) / (R * T2)

Finally, with n2, we can calculate the new volume (V2) of oxygen gas produced by multiplying n2 by the molar volume of an ideal gas at standard temperature and pressure (STP):

V2 = n2 * (22.4 L/mol)

Of course! To set up this problem, we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law is given by the formula:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

In this case, we want to find the amount of oxygen gas produced at a different temperature while keeping all the other variables the same. Since the number of moles is proportional to the volume and the ratio of moles to volume remains constant, we can set up the following equation:

(V1 / V2) = (T1 / T2)

where V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.

Given that 12 mL of oxygen gas is produced in 100 seconds at 298 K, we can assume that the initial volume, V1, is 12 mL. Now, let's solve for V2, the final volume, using the equation above.

(V1 / V2) = (T1 / T2)
(12 mL / V2) = (298 K / 308 K)

Now, let's rearrange the equation to solve for V2:

V2 = V1 * (T2 / T1)
V2 = 12 mL * (308 K / 298 K)

Using this equation, you can calculate the final volume, V2. After finding the final volume, you can compare it to the initial volume, V1, to determine the amount of oxygen gas produced in the new scenario.

Remember to convert the volumes to the same units (e.g., mL to L) and the temperatures to Kelvin before performing the calculations.

Once you have the final volume, you can multiply it by the molar volume of oxygen gas at the given conditions (usually around 22.4 L/mol, but you can check your specific conditions) to find the number of moles of oxygen gas produced. From there, you can convert moles to grams if needed using the molar mass of oxygen (32 g/mol).

I hope this helps you set up the problem! Let me know if you have any further questions.

I think the secret word here is "estimate". I see that the temperature is changing by 10 degrees. There is an estimate in reaction rates that a reaction doubles for every 10 degrees change. So 2 x 12 mL = ?