Variation of log10 K with (1/T) is shown by the following graph in which straight line is at 45Â°, hence Δ Ho is:

Question

Variation of log10 K with (1/T) is shown by the following graph in which straight line is at 45Â°, hence Δ Ho is: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/a586b3f757d26117edd3c5623abd9abc-.jpeg" /> (A) +4.606 cal (B) -4.606 cal (C) 2 cal (D) - 2 cal

Tardigrade Admin · Accepted Answer

Slope of plot = (-Δ Ho/ 2.303R)=1 ⇒ Δ Ho =-2.303 × 2=-4.606 cal

Q. Variation of log $_{10} K$ with $\frac{1}{T}$ is shown by the following graph in which straight line is at $45°,$ hence $Δ H^{o}$ is:

$+ 4.606 c a l$

$- 4.606 c a l$

$2 c a l$

$- 2 c a l$

Slope of plot $= \frac{- Δ H ^{o}}{2.303 R} = 1$
$\Rightarrow Δ H^{o} = - 2.303 \times 2 = - 4.606$ cal

Q. Variation of log10​K with T1​ is shown by the following graph in which straight line is at 45°, hence ΔHo is:

+4.606cal

−4.606cal

2cal

−2cal