Respuesta :
Answer : The partial pressure of [tex]O_2[/tex] is, 222.93 torr
Explanation :
Half-life = 2.81 hr = 168.6 min
First we have to calculate the rate constant, we use the formula :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{168.6min}[/tex]
[tex]k=4.11\times 10^{-3}min^{-1}[/tex]
Now we have to calculate the partial pressure of [tex]O_2[/tex]
The balanced chemical reaction is:
[tex]2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)[/tex]
Initial pressure 760 0 0
At eqm. (760-2x) 4x x
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{P_o}{P_t}[/tex]
where,
k = rate constant
t = time passed by the sample = 215 min
a = initial pressure of [tex]N_2O_5[/tex] = 760 torr
a - x = pressure of [tex]N_2O_5[/tex] at equilibrium = (760-2x) torr
Now put all the given values in above equation, we get:
[tex]215=\frac{2.303}{4.11\times 10^{-3}}\log\frac{760}{760-2x}[/tex]
[tex]x=222.93torr[/tex]
The partial pressure of [tex]O_2[/tex] = x = 222.93 torr
We have that the partial pressure of O2 is mathematically given as
x = 216.7 torr
From the question we are told
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Partial pressure
Generally the equation for the rate constant is mathematically given as
k = 0.693/ 2.81
k=0.247 h-1
Where
t = 205 min
t= 205 / 60
t=3.42
With
[tex]k = 1/t*ln\frac{Po}{Pt}\\\\0.247 = 1/3.42 * ln\frac{760}{Pt}\\\\Pt = 326.6 torr[/tex]
Generally the Partial pressure of Oxygen
- x = 216.7 torr
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