Explanation
Order of reaction with respect to $$O_2: first$$
Order of reaction with respect to $$NO: second$$
$$Rate = k [NO]^2[O]$$
Volume is decreased to half, then concentration of each reactants gets doubled.
$$Rate \ New= k[2NO]^2[2O]$$
$$\therefore$$ Rate of reaction will increase to eight times of it’s initial value.
By using arrhenius equation, $$ \implies \log k = \frac{-E_a}{2.303\space RT} + \log A$$
We get $$ \implies \log A = \log(1.8 \times 10^{-5}) + \frac{94140}{2.303 \times 8.314 \times 313 }$$ $$ \implies (\log 1.8) - 5 + 15.7082)$$ $$\implies 0.2553 - 5 + 15.7082 = 10.9635$$
$$ \therefore \log A = (10.9634) = 9.914 \times 10^{10} $$
We get $$ \implies \log A = \log(2.418 \times 10^{-5}) + \frac{179900}{2.303 \times 8.314 \times 546 }$$ $$ \implies (\log 2.418) - 5 + 15.7082)$$ $$\implies 0.3834 - 5 + 17.209 = 12.6$$
$$ \therefore \log A = (12.6) = 3.9 \times 10^{12} $$
In a first-order reaction the concentration of the reactant is decreased from $$1.0{\text{ M to 0}}{\text{.25 M }}$$ in 20 min.The rate constant of the reaction would be:
By first order kinetic rate constant,
$$ k = \cfrac{2.303}{t} \log(\cfrac{a}{a-x})$$
$$a = 0.1 \space M$$
$$ (a-x) = 0.05 \space M$$
$$ t = 20 \space min$$
$$ \implies k = \cfrac{2.303}{20} \log(\cfrac{0.1}{0.05})$$
$$ \implies k = 0.0347 \space min^{-1}$$
$$ Rate= \cfrac{dx}{dt} = k[A]^1$$
$$\implies 0.0347 \times 0.02$$
$$ \implies 6.94 \times 10^{-4} \times M \space min^{-1}$$
For which order reaction a straight line is obtained along with $$x - axis$$ by plotting a graph between half-life $$\left( {{t_{1/2}}} \right)$$ and initial concentration $$'a'$$.
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