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The instantaneous rate of an elementary ...

The instantaneous rate of an elementary chamical reactkon `aA+bBhArr cC+dD` can be given by rate `=K_(f)[A]^(a)[B]^(b)-K_(b)[C]^(c)[D]^(d)` where `K_(f)` and `K_(b)` are rate constants for forward and backward reactions respectively for the reversible reaction. If the reaction is an irreversible one, the rate is expressed as, rate `=K[A]^(a)[B]^(b)` where K is rate contant for the given irreversible rate of disappearance of A is a/b times the rate of disappearance of B. The variation of rate constant K with temperature is expressed in terms of Arrhenius equation: `K=Ae^(-E_(a)//RT)` whereas the ratio `(K_(f))/(K_(b))` is expressed in terms of van't Hoff isochore: `(K_(f))/(K_(b))=Ae^(-DeltaH//RT)`, where `E_(a)` and `DeltaH` are energy of activation and heat of reaction respectively.
The variation of rate constant K and `(K_(f))/(K_(b))` with temperature shows the following effects:
For endothrmic reaction when T increases then K increases and `(K_(f))/(K_(b))` also increases.
(ii) For endothemic reaction when T decreases then K decreases and `(K_(f))/(K_(b))` also decreases.
(iii) For exothermic when T increases then K and `(K_(f))/(K_(b))` both increases.
(iv) For exothermic reaction when T decreases then K increases and `(K_(f))/(K_(b))` decrease.
(v) For exothermic reaction when T increases thenK and `(K_(f))/(K_(b))` both decrease.

A

I,ii

B

iii,v

C

ii,iii

D

ii,iii,v

Text Solution

Verified by Experts

The correct Answer is:
A

`K_(C)=((K_(f))/(K_(b)))`, rate constant `propK`
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Knowledge Check

  • The instantaneous rate of an elementary chamical reactkon aA+bBhArr cC+dD can be given by rate =K_(f)[A]^(a)[B]^(b)-K_(b)[C]^(c)[D]^(d) where K_(f) and K_(b) are rate constants for forward and backward reactions respectively for the reversible reaction. If the reaction is an irreversible one, the rate is expressed as, rate =K[A]^(a)[B]^(b) where K is rate contant for the given irreversible rate of disappearance of A is a/b times the rate of disappearance of B. The variation of rate constant K with temperature is expressed in terms of Arrhenius equation: K=Ae^(-E_(a)//RT) whereas the ratio (K_(f))/(K_(b)) is expressed in terms of van't Hoff isochore: (K_(f))/(K_(b))=Ae^(-DeltaH//RT) , where E_(a) and DeltaH are energy of activation and heat of reaction respectively. For an elementary reaction aAto product, the graph plotted log([-d[A]])/(dt) vs log[A]_(t) gives a straight line with intercept equal to 0.6 and showing an angle of 45^(@) then

    A
    rate constant =4`"time"^(-1)` and a=1
    B
    rate constant `=4"mol"L^(-1)t^(-1)` and a=1
    C
    rate constant =1.99`"time"^(-1)` and a=1
    D
    rate constant `=1.99"mol"^(-1)L^(-1)` and a=2
  • The instantaneous rate of an elementary chamical reactkon aA+bBhArr cC+dD can be given by rate =K_(f)[A]^(a)[B]^(b)-K_(b)[C]^(c)[D]^(d) where K_(f) and K_(b) are rate constants for forward and backward reactions respectively for the reversible reaction. If the reaction is an irreversible one, the rate is expressed as, rate =K[A]^(a)[B]^(b) where K is rate contant for the given irreversible rate of disappearance of A is a/b times the rate of disappearance of B. The variation of rate constant K with temperature is expressed in terms of Arrhenius equation: K=Ae^(-E_(a)//RT) whereas the ratio (K_(f))/(K_(b)) is expressed in terms of van't Hoff isochore: (K_(f))/(K_(b))=Ae^(-DeltaH//RT) , where E_(a) and DeltaH are energy of activation and heat of reaction respectively. For a gaseous phase -I order reaction A(g)toB(g)+2C(g) (rate constant K=10^(-2)"time"^(-1) ) in a closed vesel of 2 litre containing 5 mole of A(g) at 27^(@)C which of the following is correct?

    A
    Rate of appearance of C (g) is `5xx10^(-2),"mol"L^(-1)t^(-1)`
    B
    Rate of disappearance of A(g) is `6.15xx10^(-3)'atm"t^(-1)`
    C
    Rate of disappearance of A (g) is `5.0xx10^(-2)"mol"t^(-1)`
    D
    Rate of appearance of B (g) is `5xx10^(-2)"moL"^(-1)t^(-1)`
  • For a reaction (1)/(2)A to 2B rate of disappearance of A is related to rate of apperarance of B by the expression

    A
    `(-d[A])/(dt)=(1)/(2)(d[B])/(dt)`
    B
    `(-d[A])/(dt)=4(d[B])/(dt)`
    C
    `(-d[A])/(dt)=(1)/(4)(d[B])/(dt)`
    D
    `(-d[A])/(dt)=(d[B])/(dt)`
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