Beta particle in the reaction
Answer: 1.87 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 = 2.50 atm
= final pressure of gas = ?
= initial volume of gas = 26.4 ml
= final volume of gas = 36.2 ml
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
The new pressure is 1.87 atm by using combined gas law.
the calculated value is Ea is 18.2 KJ and A is 12.27.
According to the exponential part in the Arrhenius equation, a reaction's rate constant rises exponentially as the activation energy falls. The rate also grows exponentially because the rate of a reaction is precisely proportional to its rate constant.
At 500K, K=0.02s−1
At 700K, k=0.07s −1
The Arrhenius equation can be used to calculate Ea and A.
RT=k=Ae Ea
lnk=lnA+(RT−Ea)
At 500 K,
ln0.02=lnA+500R−Ea
500R Ea (1) At 700K lnA=ln (0.02) + 500R
lnA = ln (0.07) + 700REa (2)
Adding (1) to (2)
700REa100R1[5Ea-7Ea] = 0.02) +500REa=0.07) +700REa.
=ln [0.02/0 .07]
Ea= 2/35×100×8.314×1.2528
Ea =18227.6J
Ea =18.2KJ
Changing the value of E an in (1),
lnA=0.02) + 500×8.314/18227.6
= (−3.9120) +4.3848
lnA=0.4728
logA=1.0889
A=antilog (1.0889)
A=12.27
Consequently, Ea is 18.2 KJ and A is 12.27.
Learn more about Arrhenius equation here-
brainly.com/question/12907018
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Answer:
The ratio of f at the higher temperature to f at the lower temperature is 5.356
Explanation:
Given;
activation energy, Ea = 185 kJ/mol = 185,000 J/mol
final temperature, T₂ = 525 K
initial temperature, T₁ = 505 k
Apply Arrhenius equation;
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D)
Where;
is the ratio of f at the higher temperature to f at the lower temperature
R is gas constant = 8.314 J/mole.K
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7B185%2C000%7D%7B2.303%20%5Ctimes%208.314%7D%20%5B%5Cfrac%7B1%7D%7B505%7D%20-%5Cfrac%7B1%7D%7B525%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%200.7289%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%2010%5E%7B0.7289%7D%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%205.356)
Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356
