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3 years ago

7

(6 points) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m

(t) of the remaining isotope Alice owns after t years.

1 answer:

Cloud [144]3 years ago

6 0

<u>Answer:</u> The equation to calculate the mass of remaining isotope is $[A]=\frac{20}{10^{-0.217t}}$

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = $\ln 4=1.386yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{1.386yrs}=0.5yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.5yr^{-1}$

t = time taken for decay process

$[A_o]$ = initial amount of the sample = 20 grams

[A] = amount left after decay process = ? grams

Putting values in above equation, we get:

$0.5=\frac{2.303}{t}\log\frac{20}{[A]}$

$[A]=\frac{20}{10^{-0.217t}}$

Hence, the equation to calculate the mass of remaining isotope is $[A]=\frac{20}{10^{-0.217t}}$

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Answer:

(a) Rate at which $NO_2$ is formed is 0.050 M/s

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

Explanation:

The given reaction is:-

$2NO+O_2\rightarrow 2NO_2$

The expression for rate can be written as:-

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Given that:- $\frac{d[NO]}{dt}=-0.050\ M/s$ (Negative sign shows consumption)

$-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

$-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}$

$-(-0.050\ M/s)=\frac{d[NO_2]}{dt}$

$\frac{d[NO_2]}{dt}=0.050\ M/s$

(a) Rate at which $NO_2$ is formed is 0.050 M/s

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}$

$-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}$

$\frac{d[O_2]}{dt}=0.0250\ M/s$

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

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At 85°C, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate the composition of a mixture of A and B that b

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Answer:

Composition of the mixture:

$x_{A} =0.652=65.2$ %

$x_{B} =0.348=34.8$ %

Composition of the vapor mixture:

$y_{A} =0.809=80.9$ %

$y_{B} =0.191=19.1$ %

Explanation:

If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with $x_{A}$ and $x_{B}$ molar fractions can be calculated as:

$P_{vap}=x_{A}P_{A}+x_{B}P_{B}$

Where $P_{A}$ and $P_{B}$ are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when $P_{vap}=P$ . When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:

$0.60*760=P_{vap}=x_{A}P_{A}+x_{B}P_{B}\\456=x_{A}P_{A}+(1-x_{A})P_{B}\\456=x_{A}*(P_{A}-P_{B})+P_{B}\\\frac{456-P_{B}}{P_{A}-P_{B}}=x_{A}\\\\\frac{456-250}{566-250}=x_{A}=0.652$

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