Answer : The partial pressure of 
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 :
Now we have to calculate the partial pressure of
The balanced chemical reaction is:
Initial pressure 760 0 0
At eqm. (760-2x) 4x x
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = time passed by the sample = 215 min
a = initial pressure of 
= 760 torr
a - x = pressure of at equilibrium = (760-2x) torr
Now put all the given values in above equation, we get:
The partial pressure of = x = 222.93 torr
I think the answer is teh first choice. Cockroaches can live through atomic radiation!
Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:
Rate factor in the presence of catalyst:
Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:
![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;
Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
