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Which In K vs 1/T plot is correct for an...

Which In K vs 1/T plot is correct for an equilibrium that shits toward reactants at higher temperatures?

A

B

C

D

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The correct Answer is:
C
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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

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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 If for a particular reversible reaction K_(C)=57 abd 355^(@)C and K_(C)=69 at 450^(@)C then

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