Question

For the dissociation of $PC{l}_5\left(g\right)$
$PC{l}_5\left(g\right)\rightleftharpoons PCl\left(g\right)+C{l}_2\left(g\right)$
Slope of the linear curve is such that $\theta =ta{n}^{-1}(-2.1)$

Thus, $△H^o$ is

• A. $28.72Jmo{l}^{-1}$
• B. $-28.72Jmo{l}^{-1}$
• C. $40.2Jmo{l}^{-1}$
• D. $-12.47Jmo{l}^{-1}$

Answer 1

The correct option is C $40.2Jmo{l}^{-1}$

Van't Hoff equation is
$lo{g}_{10}K=-\frac{△H^o}{2.303RT}+logC$
where C is a constant.
$\text{Graph between}lo{g}_{10}Kand\frac{1}{T}\text{is linear}.$
On comparing it to
$y=mx+c$
slope, $m=-\frac{△H^o}{2.303R}.....1$
but, $\text{slope m also equals},tan\theta$
since $\theta =ta{n}^{-1}(-2.1)$
So,
$slope,m=tan\theta =-2.1......2$
comparing equation 1 and 2 , we get
$-\frac{△H^o}{2.303R}=tan\theta$

putting the values we get,
$\therefore △H^o=2.303R\times 2.1$
$=2.303\times 8.314\times 2.1=40.2Jmo{l}^{-1}.$