A si sample doped with 10^6 atm/cm^3 what is the equilibrium constant at 300k
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explain given statement

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Van't Hoff equation: logK2K1=ΔH2.303×R[T2−T1T1T2]logK2K1=ΔA2.303×R[T2−T1T1T2]
When ΔA=−veΔH=−ve, then the reaction is exothermic
When ΔA=+veΔH=+ve, then the reaction is endothermic$
Answer: greater than that of forward reaction
According to the Van't Hoff equation,
logK2K1=ΔH2.303×R[T2−T1T1T2]logK2K1=ΔH2.303×R[T2−T1T1T2]
By putting the values provided,
ΔH=2.303×RT1T2(T2−T1)logK2K1ΔH=2.303×RT1T2(T2−T1)logK2K1
=2.303×8.314×300×400400−300log1025=2.303×8.314×300×400400−300log1025
=2.303×8.314×300×400100log1025=2.303×8.314×300×400100log1025 < 0
As ΔHΔH is -ve, the reaction is exothermic

A)
Need homework help? Click here.
Need revision notes? Click here.
Toolbox:
Van't Hoff equation: logK2K1=ΔH2.303×R[T2−T1T1T2]logK2K1=ΔA2.303×R[T2−T1T1T2]
When ΔA=−veΔH=−ve, then the reaction is exothermic
When ΔA=+veΔH=+ve, then the reaction is endothermic$
Answer: greater than that of forward reaction
According to the Van't Hoff equation,
logK2K1=ΔH2.303×R[T2−T1T1T2]logK2K1=ΔH2.303×R[T2−T1T1T2]
By putting the values provided,
ΔH=2.303×RT1T2(T2−T1)logK2K1ΔH=2.303×RT1T2(T2−T1)logK2K1
=2.303×8.314×300×400400−300log1025=2.303×8.314×300×400400−300log1025
=2.303×8.314×300×400100log1025=2.303×8.314×300×400100log1025 < 0
As ΔHΔH is -ve, the reaction is exothermic
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