The reaction A to products is first order in A.
1.) If 2.00 g A is allowed to decompose for 42 min, the mass of A remaining undecomposed is found to be 0.50 g. What is the half-life, t_{1/2}, of this reaction?
2.)Starting with 2.00 g A, what is the mass of A remaining undecomposed after 1.00 h?
To answer these questions, we need to use the concept of first-order reactions. First-order reactions follow a rate equation of the form: rate = k[A], where [A] represents the concentration of A and k is the rate constant.
1.) To find the half-life, t_{1/2}, of the reaction, we can use the formula: t_{1/2} = ln(2) / k.
To determine k, we can use the given information that when 2.00 g of A decomposes for 42 minutes, the remaining mass of A is 0.50 g.
First, we need to convert the masses of A to concentrations by dividing them by the molar mass of A (assuming A is a compound with a known molar mass). Let's call the molar mass of A "M".
Concentration of A at t=0: [A]0 = 2.00 g / M
Concentration of A at t=42 min: [A]t = 0.50 g / M
Since the reaction is first-order, we can set up the following equation:
ln([A]0 / [A]t) = kt
Rearranging the equation, we have:
ln(2.00 g / 0.50 g) = k * 42 min
Simplifying, ln(4) = k * 42 min
Now, we can solve for k:
k = ln(4) / 42 min
With the value of k, we can calculate the half-life:
t_{1/2} = ln(2) / k
Substituting the value of k, we get:
t_{1/2} = ln(2) / (ln(4) / 42 min)
Simplifying further, we have:
t_{1/2} = 42 min * ln(2) / ln(4)
2.) To find the mass of A remaining undecomposed after 1.00 hour, we can use the formula: [A]t = [A]0 * e^(-kt).
Let's use the concentration of A at t=0 as [A]0 and the desired time as t:
[A]t = [A]0 * e^(-kt)
We already have the value of [A]0 as 2.00 g / M. To find [A]t after 1.00 hour, we need to calculate the value of k using the same method as in question 1.
Once we have the value of k, we can substitute the values into the equation to find the concentration [A]t at t=1.00 hour. Finally, multiply [A]t by M to get the mass of A remaining undecomposed.
1).
ln(No/N) = kt
ln(2/0.5) = k*42
solve for k, then substitute for k in th following:
k = 0.693/t1/2 and solv for
t1/2
For 2).
ln(2/N) = k*60
solve for N. Use k from the first part.