Question

For the reaction: $PC{l}_5\left(g\right)\to PC{l}_3\left(g\right)+C{l}_2\left(g\right)$:

• A. $\Delta H=\Delta E$
• B. $\Delta H>\Delta E$
• C. $\Delta H<\Delta E$
• D. Can't predicted

Answer 1

The correct option is B $\Delta H>\Delta E$

Solution:- (B) $\Delta H>\Delta E$

${P{Cl}_5}_{\left(g\right)}⟶{P{Cl}_3}_{\left(g\right)}+{{Cl}_2}_{\left(g\right)}$

$\Delta n_g=n_P-n_R=(1+1)-1=1=+ve$

As we know that,

$\Delta H=\Delta E+\Delta n_{g}RT$

$\because \Delta n_g=+ve$

$\therefore \Delta H>\Delta E$