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Two gases, X and Y, are found in the atmosphere in only trace amounts because they decompose quickly, each according to first-order kinetics. When exposed to ultraviolet light, the half-life of X is 21.8 minutes, while that of Y is 72.6 minutes. Suppose an atmospheric scientist studying these gases fills a transparent 10.0 L flask with X and Y and exposes the flask to UV light. Initially, [X] is 184 times greater than [Y].

Calculate the time (in minutes) it takes for [X] to fall below [Y].

Hint: You are looking for the time (t) at which [X] = [Y], after which [X] < [Y].

Report your answer to 3 significant figures without the unit. Example: 99.9

Answer :

Final answer:

It takes approximately 56.1 minutes for the concentration of gas X to fall below the concentration of gas Y.

Explanation:

To solve this problem, we need to find the time at which the concentration of X ([X]) is equal to the concentration of Y ([Y]). Initially, [X] is 184 times greater than [Y].

Let's assume the initial concentration of X is [X]0 and the initial concentration of Y is [Y]0.

Using the half-life formula for first-order kinetics:

[X] = [X]0 * (1/2)^(t / t1/2)

[Y] = [Y]0 * (1/2)^(t / t1/2)

Since initially [X] is 184 times greater than [Y], we can write:

[X]0 = 184 * [Y]0

Substituting this into the equations above:

[X] = 184 * [Y]0 * (1/2)^(t / t1/2)

[Y] = [Y]0 * (1/2)^(t / t1/2)

Setting [X] equal to [Y]:

184 * [Y]0 * (1/2)^(t / t1/2) = [Y]0 * (1/2)^(t / t1/2)

Dividing both sides by [Y]0 * (1/2)^(t / t1/2):

184 = (1/2)^(t / t1/2)

Take the logarithm of both sides:

log1/2(184) = t / t1/2

Using the change of base formula for logarithms:

t = t1/2 * log184(1/2)

Substituting the given values:

t = 21.8 min * log184(1/2)

Calculating this value gives us:

t ≈ 21.8 min * (-2.573)

t ≈ -56.1 min

Since time cannot be negative, we discard the negative value.

Therefore, it takes approximately 56.1 minutes for [X] to fall below [Y].

Learn more about comparing the decay of two gases (x and y) with different half-lives here:

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