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 :

P - 0, Q - 1, R - 2, S - 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P - 3, Q - 2, R - 1, S - \frac{1}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P - 1, Q - 2, R - 3, S - 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P - 0, Q - 3, R - 2, S - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $P - 3, Q - 2, R - 1, S - \frac{1}{2}$
Given that,

$R a t e = k [A_{0}]^{n}$ ----- 1

Taking log on both sides, we get

$l o g r a t e = n l o g [A]_{o} + l o g k$

As $n$ increases the slope of graph increases.

Hence, $P - 3, Q - 2, R - 1, S - \frac{1}{2}$

Suggest Corrections

Similar questions

Q. 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:

If log S = log C + Z log A is similar to the linear equation: y = c+mx, then match the following:
$\begin{matrix} Column 1 & Column 2 1. y & p. Z 2. m & q. log A 3. x & r. log C 4. c & s. log S \end{matrix}$

To plot the graphs of the following equations:
Equation 1: $3 x + 2 y = 1$
Equation 2: $y - 2 x = 2$
Fill in the values (p, q, r, s) in the given tables.
Equation 1:
$\begin{matrix} x & 0 & 5 y & p & q \end{matrix}$
Equation 2:
$\begin{matrix} x & r & s y & 0 & 5 \end{matrix}$

Q. Let A = [p, q, r, s] and B = [1, 2, 3]. Which of the following relations from A to B is not a function?

(a) R₁ = [(p, 1), (q, 2), (r, 1), (s, 2)]
(b) R₂ = [(p, 1), (q, 1), (r, 1), (s, 1)]
(c) R₃ = [(p, 1), (q, 2), (p, 2), (s, 3)
(d) R₄ = [(p, 2), (q, 3), (r, 2), (s, 2)].

$\begin{matrix} L i s t - I & L i s t - I I P . & I f_{α} = \frac{π}{7} t h e n \frac{1}{c o s α} + \frac{2 c o s α}{c o s 2 α} = & 1. & 2 Q . & u n d e r s e t x \to \infty L t [(x - 1) (x - 2) (x + 3) (x + 5) (x + 10)]^{\frac{1}{5}} - x = & 2. & 3 R . & \int_{- 2}^{2} \frac{3 x^{2} d x}{1 + e^{x}} = & 3. & 4 S . & L e t f (x) = x^{3} + x^{2} + 2 x - 1. T h e m i n i m u m i n t e g r a l v a l u e o f x i f x s a t i s f i e s f (f (x)) > f (2 x + 1) i s & 4. & 8 \end{matrix}$

Join BYJU'S Learning Program

For nth order reaction(dxdt)=Rate=k[A0]nGraph between log (rate) against log[A0] is of the typeLines P,Q,R,S are for the order :

The correct option is B P−3,Q−2,R−1,S−12Given that,Rate=k[A0]n ----- 1Taking log on both sides, we getlog rate=n log[A]o+log kAs n increases the slope of graph increases. Hence, P−3,Q−2,R−1,S−12

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 :

The correct option is B $P - 3, Q - 2, R - 1, S - \frac{1}{2}$
Given that,

$R a t e = k [A_{0}]^{n}$ ----- 1

Taking log on both sides, we get

$l o g r a t e = n l o g [A]_{o} + l o g k$

As $n$ increases the slope of graph increases.

Hence, $P - 3, Q - 2, R - 1, S - \frac{1}{2}$