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(1) 0.003 kg hr–1

(2) 0.007 kg hr–1

(3) 0.006 kg hr–1

(4) 0.00017 kg hr–1

C2H5 ---> C2H5 +HI is1.6*10 POWER -5 AT 600K AND 6.36 -10 POWER -3 AT 700KCALCULATE THE ENERGY OF ACTIVATION FOR THIS REACTION_{325}= 0.12 kcal , E_{a}(b) 0.02kcal22. In dilute aqueous H

_{2}SO_{4}, the complex diaquoudioxalatoferrate (II) is oxidized by MnO_{4}^{-}. For this reaction, the ratio of the rate of change of [H^{+}] to the rate of change of [MnO_{4}^{-}] isWooden article and freshly cut tree show activity of 7.6 and and 15.2 min

^{-1}gm^{-1}of carbon (t_{1/2}= 5760 years) respectively. The age of article in years, is $\left(A\right)5760\left(B\right)5760\times \left(\frac{15.2}{7.6}\right)\left(C\right)5760\times \left(\frac{7.6}{15.2}\right)\left(D\right)5760\times \left(15.2-7.6\right)$

A(s)

^{ H2O}------ 2B(s) + C(s)If initially two moles of A is dissolved in 360g of H

_{2}O and left for decomposition at constant temperature (25 deg) Then P_{S}in the given table is ( assuming A,B and C are miscible in water)S.N. Time The vapour pressure of solution

1. 12 hr 20mm Hg

2 . 80hr P

_{S}Vapour pressure of H

_{2}O at 25 deg is 24mm Hg(Log2 = 0.30)a) e

^{-k/x}b) x/k

c) k

d) e

^{k/x}11. Statement -1: The half life of reaction with rate law, rate of reaction = $\frac{{\mathrm{k}}_{1}\left[\mathrm{A}\right]}{1+{\mathrm{k}}_{2}\left[\mathrm{A}\right]}$ is depend on initial concentration of reactant A at very high concentration.

Statement -2 : The reaction is first order at very low concentration of A.

(A) Statement -1 is true, statement-2 is true and Statement-2 is correct explanation for Statement-1.

(B) Statement -1 is true, Statement-2 true of Statement-2 is NOT correct explanation for Statement -1.

(C) Statement -1 is true, Statement-2 is false

(D) Statement -1 is false, Statement -2 is true.

Q12. For the reaction A $\to $B, the rate law expression is – $\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}$= k [A]

^{1/2}. If initial concentration of A is [A]_{0}, then(A) The integerated rate expression is k = $\frac{2}{\mathrm{t}}$ $\left({\mathrm{A}}_{{}_{0}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\mathrm{A}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$

(B) The graph of $\sqrt{\mathrm{A}}$ Vs t will be

(C) The half life period t

_{1}_{/2}= $\frac{\mathrm{K}}{2{\left[\mathrm{A}\right]}_{0}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$(D) The time taken for 75% completion of reaction t

_{3/4}= $\frac{\sqrt{{\left[\mathrm{A}\right]}_{0}}}{\mathrm{K}}$57. The reaction $2\mathrm{NO}+{\mathrm{Br}}_{2}\to 2\mathrm{NOBr}$, is supposed to follow the following mechanism.

$\left(i\right)NO+B{r}_{2}\stackrel{fast}{\rightleftharpoons}NOB{r}_{2}\phantom{\rule{0ex}{0ex}}\left(ii\right)NOB{r}_{2}+NO\stackrel{slow}{\to}2NOBr$

Suggest the rate of law expression.

What is the value of of frequency factor?