For the reaction PCl_{5(g)} to PCl_{3(g)} + C | vedclass

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(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \De

Vedclass · Accepted Answer

$\Delta H > \Delta E$

Vedclass · Answer

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.For the reaction $PCl_{5(g)} 	o PCl_{3(g)} + Cl_{2(g)}$,the change in the number of gaseous moles $(\Delta n_g)$ is calculated as: $\Delta n_g = (	ext{moles of gaseous products}) - (	ext{moles of gaseous reactants}) = (1 + 1) - 1 = 1$.Since $\Delta n_g = 1$,which is greater than $0$,the term $\Delta n_g RT$ is positive.Therefore,$\Delta H = \Delta E + RT$,which implies $\Delta H > \Delta E$.

For the reaction $PCl_{5(g)} \to PCl_{3(g)} + Cl_{2(g)}$,which of the following relations is correct?

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