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Standard XII

Physics

Distance Time Graph

For nth order...

Question

For $n^{t h}$ order reaction,
$(\frac{d x}{d t}) = r a t e = k [A]_{0}^{n}$
If graph between log (rate) against log $[A]_{0}$ is of the following type, then lines P, Q, R and S are of the order:

P Q R S
0 1 2 3

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P Q R S
3 2 1 0

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P Q R S
1 2 3 0

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P Q R S
0 3 2 1

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Solution

The correct option is B P Q R S
3 2 1 0
The given rate equation is,
$(\frac{d x}{d t}) = k [A]_{o}^{n}$

So,
$l o g (\frac{d x}{d t}) = n l o g [A]_{0} + l o g k$

Thus, n = slope = order
-
Hence greater the order, larger is the slope .

From the graph,

thus, P = 3, Q = 2 , R = 1, S = 0

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Similar questions

Q. For

n^{t h}

order reaction

(\frac{d x}{d t}) = R a t e = k [A_{0}]^{n}

Graph between log (rate) against

l o g [A_{0}]

is of the type
Lines

P, Q, R, S

are for the order :

Q. For the reaction,

a P \to b Q

Given that:

l o g [(R O D)_{P}] = l o g [(R O A)_{Q}] + 1

where

(R O D)_{P}

is rate of disappearance of 'P',

\frac{- d [P]}{d t}

and

(R O A)_{Q}

is rate of appearance of 'Q',

\frac{+ d [Q]}{d t}

. Then

a : b

will be:

Q. Graph between log K and

\frac{1}{T}

[Where K is rate constant

(S^{- 1})

and T is temperature (K)] is a straight line with

O X = 5, θ = t a n^{- 1} [- \frac{1}{2.303}]

Hence

E_{a}

and log A respectively will be:

Q. In the reaction,

A + B ⟶ C + D

, the rate

(\frac{d x}{d t})

when plotted against time '

t

' gives a straight line parallel to time axis and at some time

^{'} t^{'}

\frac{d x}{d t} = k

. The order and rate of reaction will be:

Q. If

p^{t h}, q^{t h}, r^{t h}

terms of a

G . P

. are the positive numbers

a, b, c

, then angle between the vectors

log a^{3}^i + log b^{3}^j + log c^{3}^k

and

(q - r)^i + (r - p)^j + (p - q)^k

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For nth order reaction, (dxdt)=rate=k[A]n0 If graph between log (rate) against log [A]0 is of the following type, then lines P, Q, R and S are of the order:

The correct option is B P Q R S 3 2 1 0The given rate equation is, (dxdt)=k[A]no So, log(dxdt)=nlog[A]0+logk Thus, n = slope = order - Hence greater the order, larger is the slope . From the graph, thus, P = 3, Q = 2 , R = 1, S = 0

For $n^{t h}$ order reaction,
$(\frac{d x}{d t}) = r a t e = k [A]_{0}^{n}$
If graph between log (rate) against log $[A]_{0}$ is of the following type, then lines P, Q, R and S are of the order:

The correct option is B P Q R S
3 2 1 0
The given rate equation is,
$(\frac{d x}{d t}) = k [A]_{o}^{n}$

So,
$l o g (\frac{d x}{d t}) = n l o g [A]_{0} + l o g k$

Thus, n = slope = order
-
Hence greater the order, larger is the slope .

From the graph,

thus, P = 3, Q = 2 , R = 1, S = 0