Given: Half life period (t_1/2) = 10 mins
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
Solution:
\(k=\frac { 0.693 }{ { t }_{ \frac { 1 }{ 2 } } } =\frac { 0.693 }{ 10 } \)
= 0.693 min^-1 = 6.93 x 10^-2 min^-1
Time taken = 1 hour = 60 minutes
a = 100%
x = ?
\(k=\frac { 2.303 }{ t } log\frac { a }{ a-x } \)
\(=\frac { 2.303 }{ 60 } \) [log 100 - log(a - x)]
log 100 - log(a-x) = \(\frac { 6.93\times { 10 }^{ -2 }\times 60 }{ 2.303 } \)
log 100 - log(a-x) = 1.8060
log (a - x) = log 100 - 1.8060
= 2 - 1.8060
log (a - x) = 0.1940
a - x = Antilog of 0.1949
a - x = 1.563%

Given data: Time taken for completion of 30% of the reaction = 12 mins.
a = 100; x = 30; a - x = 70 and t =12 min
Formula: \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
Solution:
For a first order reaction,
\(\frac { 2.303 }{ 12 } \log\frac { 100 }{ 70 } \)
\(=\frac { 2.303 }{ 12 } \times 0.1549\) = 0.02972 min^-1
k = 2.97 x 10^-2 min^-1
If t = 65.33 minutes, x = ?
\(k=\frac { 2.303 }{ 65.33 } \log\frac { 100 }{ 100-x } \)
\(0.02972=\frac { 2.303 }{ 65.33 } \log\frac { 100 }{ 100-x } \)
\(\log\frac { 100 }{ 100-x } \)= \(\frac { 0.02972\times 65.33 }{ 2.303 } =0.8430\)
\(\frac { 100 }{ 100-x }\)  =Antilog of 0.8430
100 = 6.966 (100 - x)
100 = 696.6 - 6.966x
-6.966x = 100-696.6 = 596.6
\(x=\frac { 596.6 }{ 6.966 } =85.62%\)
The reaction completed in 65.33 minutes

(i) The differential rate equation will be \(-\frac { d\left[ R \right] }{ dt } \) = k [A] [B]²
(ii) If the concentration of B is increased three times, then
\(-\frac { d\left[ R \right] }{ dt } \) = k [A] [3B]² = 9.k [A] [B]²
Therefore, the rate of reaction will increase 9 times.
(iii) When the concentration of both A and B are doubled.
\(-\frac { d\left[ R \right] }{ dt } \) = k[A] [B]² 
= k [2A] [2B]²
Therefore, the rate of reaction will increase 8 times.

[R]₀ = 5g, [R] = 3g
K = 1.15 x 10^-3 s^-1
As the reaction is of first order
K = \(\frac { 2.303 }{ t } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
\(t=\frac { 2.303 }{ 1.15\times { 10 }^{ -3 }{ s }^{ -1 } } \log { \frac { 5g }{ 3g } } \)
= 200 x 10³ (log 1.667)s
= 20 x 10³ x 0.22 19 s
= 443.8 s
= 444 s (approximately)

Given data: Rate constant of a first order reaction (k) = 0.45 sec^-1
Formula: \({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
Solution: \({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } =\frac { 0.693 }{ 0.45 } \)
Half-life period = 1.54 sec.