Chemical EquilibriumHard
Question
Variation of log10 K with 
is shown by the following graph in which straight line is at 45o, hence ᐃHo is :
is shown by the following graph in which straight line is at 45o, hence ᐃHo is :Options
A.+ 4.606 cal
B.- 4.606 cal
C.2 cal
D.- 2 cal
Solution
K = A eᐃH/RT
log K = log A -
log K = log A -
log K =
+ log A.
= 1.
ᐃH = - 2.303 R = - 4.606 cal.
log K = log A -
log K = log A -
log K =
+ log A. = 1. ᐃH = - 2.303 R = - 4.606 cal.
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