In a plot of log.k vs 1/T, the slope'is ...

In a plot of log.k vs 1/T, the slope'is :

`-E_(a)//2*303`

`E_(a)//2*303 R`

`E_(a)//2*303`

`- E_(a)//2*303 R`

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The correct Answer is:

`- E_(a)//2*303 R`

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The E_a of reaction in the presence of catalyst is 4.15 KJ/mol and in absence of catalyst is 8.3KJmol^(-1) . What is the slope of the plot of lnk vs 1/T in the absence of catalyst.

The plot og log k vs 1//T helps to calculate

The E_a of reaction in the presence of catalyst is 5.25 kJ//mol in the absence of catalyst is 8.314 kJ mol^(-1) . What is the slope of the plot of Ink vs 1/T in the absence of catalyst. (R = 8.314 J k^(-1) mol^(-1))

The slope of straight line in a plot of log.[A] versus time for a first order raction is equal to -k//2.303 .

Effect of temperature on the equilibrium process analysed by using the thermodynamics From the thermodynamics reaction DeltaG^(@)=-2.30RTlogk DeltaG^(@): Standing free energy change DeltaG^(@)=DeltaH^(@)-TDeltaS^(@) …(ii) DeltaH^(@) : Standard heat of the reaction gt From eqns.(i) and(ii) -2RTlogk=DeltaH^(@)=TDeltaS^(@) DeltaS^(@) : standard entropy change implies" "logK=-(DeltaH^(@))/(2.3RT)+(DeltaS^(@))/(2.3R) Clearly, if a plot of k vs 1/T is made then it is a straight lone having slope =(-DeltaH^(@))/(2.3R) amd y intercept =(DeltaS^(@))/(2.3R) If at temperature T_(1) equilibrium constant be k_(1) and at temperature T_(2) equilibrium constant be k_(2) then : implies" "logK_(1)=-(DeltaH^(@))/(2.3RT_(1))+(DeltaS^(@))/(2.3R) ..(iv) implies" "logK_(2)=-(DeltaH^(@))/(2.3RT_(2))+(DeltaS^(@))/(2.3R) ...(v) Substracting e.q (iv) from (v), we get from the relation we can conclude that the of equilibrium constant increase in temperature for endothermic reaction eith but value of equilibrium constant decrease with the increase in temperature for exothermic reaction If statndard heat of dissociation of PCl_(5) is 230 cal then slope of the graph of log vs (1)/(T) is :

Effect of temperature on the equilibrium process analysed by using the thermodynamics From the thermodynamics reaction DeltaG^(@)=-2.30RTlogk DeltaG^(@): Standing free energy change DeltaG^(@)=DeltaH^(@)-TDeltaS^(@) …(ii) DeltaH^(@) : Standard heat of the reaction gt From eqns.(i) and(ii) -2RTlogk=DeltaH^(@)=TDeltaS^(@) DeltaS^(@) : standard entropy change implies" "logK=-(DeltaH^(@))/(2.3RT)+(DeltaS^(@))/(2.3R) Clearly, if a plot of k vs 1/T is made then it is a straight lone having slope =(-DeltaH^(@))/(2.3R) amd y intercept =(DeltaS^(@))/(2.3R) If at temperature T_(1) equilibrium constant be k_(1) and at temperature T_(2) equilibrium constant be k_(2) then : implies" "logK_(1)=-(DeltaH^(@))/(2.3RT_(1))+(DeltaS^(@))/(2.3R) ..(iv) implies" "logK_(2)=-(DeltaH^(@))/(2.3RT_(2))+(DeltaS^(@))/(2.3R) ...(v) Substracting e.q (iv) from (v), we get from the relation we can conclude that the of equilibrium constant increase in temperature for endothermic reaction eith but value of equilibrium constant decrease with the increase in temperature for exothermic reaction For exothermic reaction if DeltaS^(@)lt0 then the sketch of log k vs (1)/(T) may be

When a plot between logk and 1/T is plotted we get the graph as shown. What is the value of slope in the graph?

Rate constant k of a reaction varies with temperature according to the equation logk = constant -E_a/(2.303R)(1/T) where E_a is the energy of activation for the reaction . When a graph is plotted for log k versus 1/T , a straight line with a slope - 6670 K is obtained. Calculate the energy of activation for this reaction

Assertion (A) : k=Ae^(-E_(a)//RT) , the Arrhenius equation represents the dependence of rate constant with temperature. Reason (R ): Plot of log.k against 1//T is linear and the activation energy can be calculated with this plot.

ICSE-SAMPLE PAPER 2012-PART-II (SECTION-C)

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