Question

Variation of log10 K with $\frac { 1 } { \mathrm {~T} }$ is shown by the following graph in which straight line is at 45⁰, hence DH⁰ is :

• A.
  • 4.606 cal
• B. – 4.606 cal
• C. 2 cal
• D. – 2 cal

Answer 1

Answer: (B)

– 4.606 cal

Answer 2

log $\mathrm { K } = \log \mathrm { A } - \frac { \Delta \mathrm { H } } { 2.303 \mathrm { RT } }$.

log .

log $\mathrm { K } = \left[ - \frac { \Delta \mathrm { H } } { 2.303 \mathrm { R } } \right] \times \frac { 1 } { \mathrm {~T} } + \log \mathrm { A }$.