For the reaction-aP - bQGiven that- log-RODP- - log-ROAQ- - 1where RODP is rate of disappearance of -P-d-P-dt and ROAQ is rate of appearance of -Q-d-Q-dt- Then a-b will be-

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:

2 : 1

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

10 : 1

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

1 : 10

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

1 : 2

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

Open in App

Solution

The correct option is B $10 : 1$
Answer (b)
$a P \to b Q$
$\Rightarrow \frac{(R O D)_{P}}{a} = \frac{(R O A)_{Q}}{b}$ (By rate of reaction)
$\frac{(R O D)_{P}}{(R O A)_{Q}} = \frac{a}{b}$
By taking log on both side
$l o g \frac{(R O D)_{P}}{(R O A)_{Q}} = l o g (\frac{a}{b})$
$∴ l o g [(R O D)_{P}] = l o g [(R O A)_{Q}] + l o g (\frac{a}{b})$
$∴ l o g (\frac{a}{b}) = 1$
$\frac{a}{b} = \frac{10}{1}$
$a : b \Rightarrow 10 : 1$

Suggest Corrections

Similar questions

Q. In the following reaction,

x A \to y B

l o g [- \frac{d [A]}{d t}] = l o g [\frac{d [B]}{d t}] + 0.3

Where negative sign indicates rate of disappearance of the reactant. Thus,

x : y

is :

Q. In the following reaction :

x A \to y B

log [- \frac{d [A]}{d t}] = log [\frac{d [B]}{d t}] + log 2

where -ve sign indicates rate of disappearance of the reactant. Thus, X : Y is:

Q. For a reaction

\frac{1}{2} A \to 2 B

rate of disappearance of A is related to rate of appearance of B by the expression

Q. For the reaction

\frac{1}{2} A \to 2 B

, the rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the expression:

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 :

Join BYJU'S Learning Program

For the reaction, aP→bQ Given that: log[(ROD)P]=log[(ROA)Q]+1 where (ROD)P is rate of disappearance of 'P', −d[P]dt and (ROA)Q is rate of appearance of 'Q', +d[Q]dt. Then a:b will be:

The correct option is B 10:1Answer (b) aP→bQ ⇒(ROD)Pa=(ROA)Qb (By rate of reaction) (ROD)P(ROA)Q=ab By taking log on both side log(ROD)P(ROA)Q=log(ab) ∴log[(ROD)P]=log[(ROA)Q]+log(ab) ∴log(ab)=1 ab=101 a:b⇒10:1

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: