$Time (sec)$ Rate $(mol\, L^{-1} sec.^{-1})$
$0$ $1.60 \times 10^{-2}$
$10$ $1.60 \times 10^{-2}$
$20$ $1.60 \times 10^{-2}$
$30$ $1.60 \times 10^{-2}$
From the above data, the order of reaction is
$Time (sec)$ Rate $(mol\, L^{-1} sec.^{-1})$
$0$ $1.60 \times 10^{-2}$
$10$ $1.60 \times 10^{-2}$
$20$ $1.60 \times 10^{-2}$
$30$ $1.60 \times 10^{-2}$
From the above data, the order of reaction is
$\therefore$ order $=$ zero
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Given : $N -N =159\,kJ\,mol^{-1};\,\,H-H = 436\,kJ\,mol^{-1}$
$N \equiv N = {\text{ }}941{\text{ }}kJ{\text{ }}mo{l^{ - 1}}{\text{ }};{\text{ }}N-H{\text{ }} = {\text{ }}398{\text{ }}kJ{\text{ }}mo{l^{ - 1}}$ .....$kJ\,mol^{-1}$
$A \to B$ ${K_1} = {10^{15}}{e^{ - 25000/8.314\,T}}$
$C \to D$ ${K_2} = {10^{14}}{e^{ - 15000/8.314\,T}}$
$2 C _{( s )}+ O _{2( g )} \rightarrow 2 CO ( g )$
When $12\,g$ carbon is burnt in $48\,g$ of oxygen, the volume of carbon monoxide produced is $......\times 10^{-1}\,L$ at STP [nearest integer]
[Given : Assume $CO$ as ideal gas, Mass of $C$ is $12\,g\,mol ^{-1}$, Mass of $O$ is $16\,g\,mol ^{-1}$ and molar volume of an ideal gas at STP is $22.7\,L\, mol ^{-1}$ ]