K = 0.693/t_1/2
(a-x) ∝ (0.40 –P)atm.
K = 2.303/12 log 0.40/0.40 –P
Pcalculated = 0.1745 atm
Total pressure = 0.4 -P+P+P+P = 0.749 atm

\(\frac { dx }{ dt } =k{ \left[ A \right] }^{ 1 }{ \left[ B \right] }^{ 0 }\)
\(\\ From \ expt. \ I\)
\(2.0\times { 10 }^{ -2 }=k\left[ 0.1 \right] ,\)
\(k=2.0\times { 10 }^{ -1 }{ min }^{ -1 }\)
\(=0.2{ \ min }^{ -1 }=2\times { 10 }^{ -1 }{ min }^{ -1 }\)
\(In \ expt. \ II\)
\(\frac { dx }{ dt } =k{ \left[ A \right] }^{ 1 }\)
\(4\times { 10 }^{ -2 }=2\times { 10 }^{ -1 }\times \left[ A \right]\)
\(\Rightarrow \ \left[ A \right] =0.2 \ mol \ { L }^{ -1 }\)
\(In \ expt. \ III\)
\(Rate=\frac { dx }{ dt } =k{ \left[ A \right] }\)
\( =1.2\times 0.4\)
\(=8\times { 10 }^{ -2 }mol \ { L }^{ -1 }{ min }^{ -1 }\)
\(In \ expt.IV\)
\(2.0\times { 10 }^{ -2 }=0.2\times \left[ A \right]\)
\(\left[ A \right] =0.1mol \ { L }^{ -1 }\)

(i) \(t_{1/2}={0.693\over 2.4}\)
\(k={2.303\over 9.6}log{100\over x}\)
\(k={2.303\over 9.6}log{100\over x}\)
\(\Rightarrow\ {0.693\over 24}={2.303\over 9.6}log{100\over x}\)
\(log{100\over x}=1.2040\)
\({100\over x}=16\)
\(x={100\over 16}=16=6.25g\)
(b)\(k={2.303\over t}log{[A]_0\over [A]}\)
\(t={2.303\times2.4\over 0.693}log{5\times10^{10}\over 10^8}\)
\(t={2.303 \times2.4\over 0.693}log500\)
\(t={2.303\times2.4\times2.6990\over 0.693}\)
t = 21.526 min.

(i) \(k={0.693\over t_{1/2}}={0.693\over 2.1hr}\)
= 0.33 hr^-1
\(t_{99./.}={2.303\over k}log{[R]_0\over {1\over 100}[R]_0}\)
[90% is complete. 10% is left]
\(t_{99./.}={4.606\over k}={4.606\over 0.33hr^{-1}}\)
t_99% = 13.96 hr
(ii) Amount decomposed
= 99% of 6.2 g
\(={6.2\times{99\over 100}}=6.138g\)
1 mole of NH₂NO₂ (62 g) produces 22.4 L of N₂O at STP.
6.138 g will produce
\(={22.4\over 62}\times6.138\)
= 2.217 L of N₂O at STP.

N₂(g) + O₂(g) ⇌ 2NO(g); k = 1.0 x 10^-5
N₂(g) + O₂(g) ⇌ 2NO(g)
Initial cone. 0.80     0.20        0
Final cone. 0.80 - x 0.20-x    2x
\(k={[NO]^2\over [N_2][O_2]}\Rightarrow\ 1.0\times10^{-15}={(2x)^2\over(0.8-x)(0.2-x)}\)
Since x is very small, so 0.8 - x ≈ 0.8 and 0.2 - x ≈ 0.2, hence
\(1.0\times10^{-5}={(2x)^2\over 0.8\times0.2}\)
x = 6.3245 x 10-4 mol L^-1
[NO] = 6.3245 x 10^-4 x 2 = 1.2649 x 10^-3 mol L^-1 = 1.26 x 10-3 mol L^-1
[N₂] = 0.80 - 6.3245 x 10^-4 = 0.79936 mol L^-1 = 0.7994 mol L^-1
[O₂] = 0.2 - 6.3245 x 10^-4 = 0.19936 mol L^-1 = 0.1994 mol L^-1

\({ T }_{ 1 }=295K, \ { T }_{ 2 }=305K, \ { k }_{ 2 }=2{ k }_{ 1 } \ (given)\)
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
\(\log { 2 } = \ \frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 305-295 }{ 305\times 295 } \right) \)
\({ E }_{ a } \ = \ \frac { 19.147\times 305\times 295\times 0.3010 }{ 10 } \)
\( { E }_{ a } \ = \ 51854.8 \ J \ { mol }^{ -1 }\)
\(=51.85 \ KJ \ { mol }^{ -1 }\)

\(k={2.303\over t}log{[R]_0\over [R]}\)
\(={2.303\over 100}log{[R]_0\over {40\over 100}[R]_0}\)
[60% is complete, 40% is left]
\(k={2.303\over 100}(log5-log2)\)
\(k={2.303\over 100}(0.6990-0.3010)\)
\(k={2.303\over 100}\times0.3980\ min^{-1}\)
\(t_{90./.}={2.303\over k}log{[R]_0\over [R]}\)
[90% is complete, 10% is left]
\(t_{90./.}={2.303\over k}log{[R]_0\over {10\over 100}[R]_0}\)
\(={2.303\times100\over 2.303\times0.3980}log10\)
\(={100\over 0.3980}=251.26min\)