The air in a 2 L balloon at 0.998 atm and 34.0 °C. What will be its pressure if it is brought to a higher altitude where it now
1 answer:
Answer: 0.529 atm
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 0.998 atm
= final pressure of gas = ?
= initial volume of gas = 2 L
= final volume of gas = 3.5 L
= initial temperature of gas =
= final temperature of gas =
Now put all the given values in the above equation, we get:
Thus the pressure if it is brought to a higher altitude where it now occupies 3.5 L and is at 12.0 °C is 0.529 atm
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Answer:
The answer to the questions are as follows
Reaction B is 4426.28 times faster than reaction A
(b) Reaction B is faster.
Explanation:
To solve the question we are meant to compare both reactions to see which one is faster
The values of the given activation energies are as follows
For A
Ea = 95.00 kJ mol–1 (22.71 kcal mol–1) and
for B
Ea = 74.20 kJ mol–1 (17.73 kcal mol–1)
T is the same for both reactions and is equal to 298 k
Concentration of both reaction = 1M
The Arrhenius Law is given by
k =
Where
k = rate constant
Ea = activation energy
R = universal gas constant
T = temperature (Kelvin )
A = Arrhenius factor
Therefore
For reaction A, the rate constant k₁ is given by k₁ =
And for B the rate constant k₂ is given by k₂ =
k₁ = A×2.225×10⁻¹⁷
k₂ = A×9.850×10⁻¹⁴
As seen from the above Reaction B is faster than reaction A by (A×9.850×10⁻¹⁴)/(A×2.225×10⁻¹⁷) or 4426.28 times