Answer :
The decomposition of phosphine,[tex]PH_3[/tex], follows a rate equation of Rate = [tex]k[PH_3][/tex]. The half-life of [tex]PH_3[/tex] is 37.9s at [tex]120^0C[/tex]. This information is used to determine the time required for three-fourths of [tex]PH_3[/tex] to decompose and the fraction remaining after 1 minute.
(a) The half-life of [tex]PH_3[/tex] is the time required for half of the original sample to decompose. In this case, the half-life is 37.9s. To find the time required for three-fourths (or 75%) of [tex]PH_3[/tex] to decompose, we need to determine how many half-lives are needed. Since each half-life reduces the amount of PH₃ by half, three-fourths is equivalent to 1.5 half-lives. Therefore, the time required for three-fourths of [tex]PH_3[/tex] to decompose is 1.5 times the half-life:
Time = 1.5 * 37.9s = 56.85s.
(b) To determine the fraction of the original sample remaining after 1 minute, we need to calculate the number of half-lives that have elapsed. The time given is 1 minute, which is equivalent to 60 seconds. Dividing the total time by the half-life gives us the number of half-lives:
A number of half-lives = 60s / 37.9s = 1.58 (approximately).
Since the reaction follows first-order kinetics, the fraction remaining is equal to 1 divided by 2 raised to the power of the number of half-lives:
Fraction remaining =[tex]1 / 2^(^1^.^5^8) = 0.405[/tex] (approximately).
Therefore, approximately 40.5% of the original sample of [tex]PH_3[/tex] remains after 1 minute.
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