A substance undergoes a first-order decomposition. After 40.0 min at 500 degrees C, only 12.5% of the original sample remains. What is the half-life of the decomposition? If the original sample weighted 243 g, how much would remain after 2 hours?
n * t1/2 = mass/(2^n)
n * t1/2 = 2 hrs = 120 mins
n = 120/13.3
n = 9.02 (n = number of half-cycles)
remaining mass = 243/(2^9.02)
= 0.467g
0.467g is remaining after 2 hrs
).5^n = 0.125
n = 3
t 1/2 = 40 mins / 3
t 1/2 = 13.3 mins
Hope that helps
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To determine the half-life of the decomposition, we can use the first-order reaction equation:
A --> B
where A represents the original substance and B represents the decomposition product.
The half-life (t1/2) is the time it takes for half of the original substance to decompose.
We can use the given information to find the half-life. We know that after 40.0 min at 500 degrees C, only 12.5% of the original sample remains. This means that 87.5% (100% - 12.5%) of the substance has decomposed.
To find the half-life, we can use the formula:
ln(Nt/N0) = -kt
where Nt is the amount of substance remaining at time t, N0 is the initial amount of substance, k is the rate constant, and t is the time.
Taking the natural logarithm of (Nt/N0) and substituting the given values, we have:
ln(0.125) = -k * 40.0 min
Now we can solve for k:
k = ln(0.125) / -40.0 min
Using this value of k, we can now find the half-life (t1/2). The half-life is given by:
t1/2 = ln(2) / k
Substituting the value of k we found, we obtain:
t1/2 = ln(2) / (ln(0.125) / -40.0 min)
Calculating this expression gives us the value of the half-life.
Now, to determine how much of the substance would remain after 2 hours (120 minutes), we can use the equation:
Nt = N0 * e^(-kt)
Substituting the values of N0 and k we found, we get:
Nt = 243 g * e^(-k * 120 min)
Calculating this expression will give us the remaining amount of the substance after 2 hours.