For a gaseous reaction- the rate is expressed interms of dPdt in place of dCdt or dndt- where C is concentration- n is number of moles and -P- ispressure of reactant- The three are related as-

Rate expression =dPdt From ideal gas equation, P=nVRT ∴ C=nV and C=PRT dCdt=ddt( nV )=1V( dndt ) ddt( PRT )=1RT( dPdt ) ...

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Standard XII

Chemistry

Factors Affecting Rate of Reaction

For a gaseous...

Question

For a gaseous reaction, the rate is expressed interms of $\frac{d P}{d t}$ in place of $\frac{d C}{d t}$ or $\frac{d n}{d t}$ , where $C$ is concentration, $n$ is number of moles and $^{'} P^{'}$ ispressure of reactant. The three are related as:

[\frac{d P}{d t}] = \frac{R T}{V} [\frac{d n}{d t}] = [\frac{d C}{d t}]

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\frac{1}{R T} [\frac{d P}{d t}] = \frac{1}{V} [\frac{d n}{d t}] = [\frac{d C}{d t}]

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[\frac{d P}{d t}] = [\frac{d n}{d t}] = [\frac{d C}{d t}]

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None of these

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Solution

The correct option is B $\frac{1}{R T} [\frac{d P}{d t}] = \frac{1}{V} [\frac{d n}{d t}] = [\frac{d C}{d t}]$
Rate expression $= \frac{d P}{d t}$
From ideal gas equation, $P = \frac{n}{V} R T$
$∴ C = \frac{n}{V}$ and $C = \frac{P}{R T}$
$\frac{d C}{d t} = \frac{d}{d t} (\frac{n}{V}) = \frac{1}{V} (\frac{d n}{d t})$
$\frac{d}{d t} (\frac{P}{R T}) = \frac{1}{R T} (\frac{d P}{d t})$

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Similar questions

Q. For a gaseous reaction, the rate is expressed interms of

\frac{d P}{d t}

in place of

\frac{d C}{d t}

\frac{d n}{d t}

, where

C

is concentration,

n

is number of moles and

^{'} P^{'}

ispressure of reactant. The three are related as:

Q. For the ideal gaseous reaction, the rate is generally expressed in terms of

\frac{d P}{d t}

instead of

\frac{d C}{d t}

\frac{d n}{d t}

(where

C = n / v

is concentration and

n

is the number of moles). What is the relation among these three expressions if

T

and

V

are constant?

Q. For gaseous reaction, the rate is often expressed in terms of dP/dt instead of dc/dt or dn/dt (where c is the concentration and n the number of mol). What is the relation among these three expressions ?

Q. Which is correct relation in between

\frac{d c}{d t}, \frac{d n}{d t}

and

\frac{d P}{d t}

, where c, n, P, represents concentration, mole and pressure terms for gaseous phase reactant

A (g) \to p r o d u c t

Q. For a gaseous reaction,

A (g) ⟶

Product, which one of the following is correct relation among

\frac{d P}{d t}, \frac{d n}{d t}

and

\frac{d c}{d t}

?
(

\frac{d P}{d t} =

Rate of reaction in

a t m

{s e c}^{- 1}

;

\frac{d c}{d t} =

Rate of reaction in molarity

{s e c}^{- 1}

;

\frac{d n}{d t} =

Rate of reaction in

m o l

{s e c}^{- 1}

)

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For a gaseous reaction, the rate is expressed interms of dPdt in place of dCdt or dndt, where C is concentration, n is number of moles and ′P′ ispressure of reactant. The three are related as:

The correct option is B 1RT[dPdt]=1V[dndt]=[dCdt]Rate expression =dPdt From ideal gas equation, P=nVRT ∴C=nV and C=PRT dCdt=ddt(nV)=1V(dndt) ddt(PRT)=1RT(dPdt)

For a gaseous reaction, the rate is expressed interms of $\frac{d P}{d t}$ in place of $\frac{d C}{d t}$ or $\frac{d n}{d t}$ , where $C$ is concentration, $n$ is number of moles and $^{'} P^{'}$ ispressure of reactant. The three are related as: