Ethane, C2H6, forms ·CH3 radicals at 700.°C in a first-order reaction, for which k = 1.98 h-1.
(a) What is the half-life for the reaction?
(b) Calculate the time needed for the amount of ethane to fall from 1.00 10-3 mol to 2.48 10-4 mol in a 500. mL reaction vessel at 700.°C.
(c) How much of a 6.87 mg sample of ethane in a 500. mL reaction vessel at 700.°C will remain after 47 mins?
See your earlier posts. Most of these are the same type question.
(a) To find the half-life of the reaction, we can use the formula for a first-order reaction:
t1/2 = 0.693 / k
where t1/2 is the half-life and k is the rate constant.
Plugging in the given value for k, we get:
t1/2 = 0.693 / (1.98 h^-1)
Calculating this expression, we find:
t1/2 ≈ 0.350 h (rounded to three decimal places)
Therefore, the half-life for the reaction is approximately 0.350 hours.
(b) To calculate the time needed for the amount of ethane to fall from 1.00 × 10^-3 mol to 2.48 × 10^-4 mol, we need to use the equation for the first-order reaction:
ln(A0/At) = kt
where A0 is the initial amount of ethane, At is the final amount of ethane, k is the rate constant, and t is the time.
Plugging in the given values, we have:
ln(1.00 × 10^-3 mol / 2.48 × 10^-4 mol) = (1.98 h^-1) × t
Simplifying:
ln(4.032) = 1.98 t
Using logarithmic properties, we find:
t ≈ ln(4.032) / 1.98
Evaluating this expression, we obtain:
t ≈ 0.304 h (rounded to three decimal places)
Therefore, it would take approximately 0.304 hours (or 18.24 minutes) for the amount of ethane to fall from 1.00 × 10^-3 mol to 2.48 × 10^-4 mol in the given reaction vessel.
(c) To find how much of a 6.87 mg sample of ethane will remain after 47 minutes, we can use the same first-order reaction equation:
ln(A0/At) = kt
But first, we need to convert the mass of ethane to moles. The molar mass of ethane (C2H6) is:
2(12.011 g/mol) + 6(1.008 g/mol) = 30.069 g/mol
Converting the sample mass:
6.87 mg = 6.87 × 10^-3 g = 6.87 × 10^-3 g / (30.069 g/mol) = 2.284 × 10^-4 mol
Now, we can plug in the values into the equation:
ln(2.284 × 10^-4 mol / At) = (1.98 h^-1) × 47 min
Converting minutes to hours:
t = 47 min / 60 min/h = 0.783 h
Rearranging the equation to solve for At:
ln(2.284 × 10^-4 mol / At) = (1.98 h^-1) × 0.783 h
ln(2.284 × 10^-4 mol / At) ≈ 1.553
Using logarithmic properties:
At ≈ (2.284 × 10^-4 mol) / e^(1.553)
Evaluating this expression, we find:
At ≈ 7.53 × 10^-5 mol
Finally, to convert the amount of ethane to mg, we multiply by the molar mass:
At = 7.53 × 10^-5 mol × (30.069 g/mol) = 2.27 × 10^-3 g
Converting to milligrams:
2.27 × 10^-3 g = 2.27 × 10^-3 g × 1000 mg/g = 2.27 mg
Therefore, approximately 2.27 mg of the 6.87 mg sample of ethane will remain after 47 minutes.
To answer these questions, we need to understand the first-order reaction kinetics and use the given rate constant (k) values.
(a) To find the half-life for the reaction, we can use the formula:
t1/2 = (0.693 / k)
Here, k is the rate constant. In the question, k is given as 1.98 h-1.
Substituting the value into the formula, we get:
t1/2 = (0.693 / 1.98)
Calculating this gives us the half-life for the reaction.
(b) To calculate the time needed for the amount of ethane to fall from 1.00 × 10-3 mol to 2.48 × 10-4 mol in a 500 mL reaction vessel at 700 °C, we can use the first-order reaction integrated rate law equation:
ln([A]t / [A]0) = -kt
Where [A]t is the final concentration, [A]0 is the initial concentration, k is the rate constant, and t is the time.
We know that [A]0 = 1.00 × 10-3 mol and [A]t = 2.48 × 10-4 mol. We also have the value for k, which is 1.98 h-1.
Rearranging the equation and substituting the values, we can solve for time (t).
(c) To determine how much of a 6.87 mg sample of ethane will remain after 47 minutes in a 500 mL reaction vessel at 700 °C, we again use the first-order reaction integrated rate law equation:
ln([A]t / [A]0) = -kt
Here, [A]0 is the initial concentration, and [A]t is the final concentration. In this case, [A]0 is given as 6.87 mg, and t is given as 47 minutes. The rate constant, k, remains the same.
Rearranging the equation and substituting the values, we can find the remaining amount of ethane in the reaction vessel.
Note: When performing calculations, be sure to use consistent units throughout, such as converting grams to moles or minutes to hours when necessary.