The reaction $v_{1} A + v_{2} B \to P r o d u c t s$ is first order w.r.t $A$ and zero order w.r.t $B$ . If the reaction is started with $[A_{0}]$ and $[B_{0}]$ the integral rate equation for the reaction would be:

l n \frac{[A_{0}]}{[A_{0}] - x} = k_{1} t

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

l n \frac{[A_{0}]}{[A_{0}] - v_{1} x} = k_{1} t

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

l n \frac{[A_{0}]}{[A_{0}] - v_{1} x} = v_{1} k_{1} t

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

l n \frac{[A_{0}]}{[A_{0}] - x} = v_{1} k_{1} t

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

Open in App

Solution

The correct option is A $l n \frac{[A_{0}]}{[A_{0}] - x} = k_{1} t$
$v_{1} A + v_{2} B ⟶$ product
Reaction is $1^{s t}$ order w.r.t A
Reaction is zero order w.r.t B.
i.e. Rate of Reaction varies only with concentration of A. It is a pseudomonomolecular reaction.
Rate constant $k_{1} = \frac{1}{t} l n \frac{[A_{o}]}{[A_{o} - x]}$
$\Rightarrow k_{1} t = l n \frac{[A_{o}]}{[A o - x]}$

Suggest Corrections

Similar questions

If [A] is the concentration of reactant A at any time t and [A_o] is the concentration at t=0, then for the first order reaction the rate equation can be written as:

Q. A reaction :
aA

\to

product
follows zero order kinetics. The concentration of A with respect to time (t) is given as:

If Avogadro's number is A $_{0}$ , the number of sulphur atoms present in 200 mL of 1N H $_{2}$ SO $_{4}$ is

Q. Show that

\frac{1}{a_{0}} + \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} + \dots + \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}} = \frac{1}{a_{0} -} \frac{a_{0} x}{a_{1} + x -} \frac{a_{1} x}{a_{2} + x -} \dots \frac{a_{n - 1} x}{a_{n} + x}

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

The reaction v1A+v2B→Products is first order w.r.t A and zero order w.r.t B. If the reaction is started with [A0] and [B0] the integral rate equation for the reaction would be:

The correct option is A ln[A0][A0]−x=k1tv1A+v2B⟶ productReaction is 1st order w.r.t AReaction is zero order w.r.t B.i.e. Rate of Reaction varies only with concentration of A. It is a pseudomonomolecular reaction.Rate constant k1=1tln[Ao][Ao−x]⇒k1t=ln[Ao][Ao−x]

The reaction $v_{1} A + v_{2} B \to P r o d u c t s$ is first order w.r.t $A$ and zero order w.r.t $B$ . If the reaction is started with $[A_{0}]$ and $[B_{0}]$ the integral rate equation for the reaction would be: