For an isothermal second order aqueous phase reaction, A → B, the ratio of the time required for 90% conversion to the time required for 45% conversion is
A. 2
B. 4
C. 11
D. 22
Answer: Option C
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A. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\text{k}} + {{\text{k}}_{\text{g}}}$$
B. $${{\text{k}}_{\text{e}}}{\text{ff}} = \frac{{{\text{k}} + {{\text{k}}_{\text{g}}}}}{2}$$
C. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\left( {{\text{k}}{{\text{k}}_{\text{g}}}} \right)^{\frac{1}{2}}}$$
D. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$
The half life period of a first order reaction is given by (where, K = rate constant. )
A. 1.5 K
B. 2.5 K
C. $$\frac{{0.693}}{{\text{K}}}$$
D. 6.93 K
Catalyst is a substance, which __________ chemical reaction.
A. Increases the speed of a
B. Decreases the speed of a
C. Can either increase or decrease the speed of a
D. Alters the value of equilibrium constant in a reversible
A. $$ \propto {\text{CA}}$$
B. $$ \propto \frac{1}{{{\text{CA}}}}$$
C. Independent of temperature
D. None of these

Here, for 2nd order reaction we have a equation:
(-rA) = -dCA/dt = KCA^2
Solving the equation we will get (1/CA - 1/CA0) = Kt
Now we have to solve for 90% conversion so we have CA = 0.1 (as 90% is consumed) and assume CA0 = 1
Similarly for 45% conversion we have CA = 0.55 and CA0 = 1
for 90% conversion we will get Kt1 = 9 and for 45% conversion we will get Kt2 = 0.82
Now just divide kt1/kt2, so we will get t1/t2 = 11
I hope you got the solution
Can I have the calculation please?
can I know how to calculate this please