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Question

Variation of log10K with 1T is shown by the following graph in which straight line is at 45o, hence ΔHo is:
698501_5904159b12424b42996e97c9160c642c.PNG

A
+4.606cal
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B
4.606cal
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C
2cal
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D
2cal
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Solution

The correct option is B 4.606cal
Given,
Angle the line makes with the horizontal=45
Therefore, slope of the line m=tan45=1 ...(i)
Now, we know that
ΔG=ΔHTΔS
2.303RT(logK)=HTΔS

Simplifying the above expression, we get
logK=ΔH2.303RT+TΔS2.303RT

Simplifying the equation further and writing it in the form of equation of a line we get,
logK=ΔH2.303R(1T)+ΔS2.303R

Comparing the above equation with the general equation of a straight line y=mx+c , we get
m=ΔH2.303R

Also, from equation (i) we get
m=1 , therefore we get 1=ΔH2.303R

Substituting the value of R=2 , we get ΔH=2.303(2)=4.606cal
Therefore, the correct answer will be option B

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