The rate constant is mathematically given as
K2=2.67sec^{-1}
<h3>What is the Arrhenius equation?</h3>
The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

Therefore
KT1= 0.0110^{-1}
T1= 21+273.15
T1= 294.15K
T2= 200
T2=200+273.15
T2= 473.15K
Ea= 35.5 Kj/Mol
Hence, in j/mol R Ea is
Ea=35.5*1000 j/mol R

K2/0.0110 =e^(5.492)
K2/0.0110 =242.74
K2= 242.74*0.0110
K2=2.67sec^{-1}
In conclusion, rate constant
K2=2.67sec^{-1}
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Answer:
Yes
Explanation:
By definition, the equilibrium constanct, Kc, for the reaction A ⇒ 2B is
= [A]^1 / [B]^2
Substitute [A] = 4 and [B] = 2 in the equation,
[A]^1 / [B]^2
= 4^1 / 2^2
= 1
= Kc
So yes the reaction is at equilibrium.
Answer:
The number of protons you welcome
Explanation:
Answer:
DUPLET RULE: The tendency of an atom to acquire an outer most shell of two electrons is called duplet rule. OCTET RULE: The tendency of an atom to acquire an outermost shell of eight electrons is called the octet rule.
Answer:
The new pressure will be 0.225 kPa.
Explanation:
Applying combined gas law:

where,
are initial pressure and volume at initial temperature
.
are final pressure and volume at initial temperature
.
We are given:



Putting values in above equation, we get:


Hence, the new pressure will be 0.225 kPa.