Question

18. For the reaction $\frac{1}{8}{\mathrm{S}}_{8\left(\mathrm{s}\right)}+\frac{3}{2}{{\mathrm{O}}_2}_{\left(\mathrm{g}\right)}\to {\mathrm{SO}}_{3\left(\mathrm{g}\right)}$, the difference of heat change at constant pressure and constant volume at 27°C will be
(1) +150 R (2) – 150 R (3) + 450 R (4) – 450 R

Answer 1

Dear Student,
Q17.According to Gibbs-Hemholtz equation
$∆H=∆U+P∆V$
The reaction is carried out in a open vessel so P will remain equal to atmospheric pressure.
$∆H=∆U+∆nRT$
For enthalpy to be equal to internal energy the second term should be equal to zero and it will be zero only when $∆n$ is equal to 0. For $∆n$ to be equal to zero , the number of moles of reactants and products should be equal which is true in case of part (4) H₂(g)+I₂(g)$\to$2HI(g)
Q18. T=27⁰ = 300K
For the reaction
$\frac{1}{8}{S}_8+\frac{3}{2}{O}_2\to S{O}_3$
we know that $∆H=∆U+P∆V$ where internal energy is the heat change at constant volume q_v and enthalpy is the heat change at constant pressure q_p
$q_p=q_v+∆nRT$
For the above reaction
$$\begin{gathered} ∆n=1-\left(\frac{3}{2}\right)[\sin cesolidS=1] \\ ∆n=-\frac{1}{2} \end{gathered}$$
$$\begin{gathered} ∆n=1-\left(\frac{3}{2}\right)[numberofmolesofsolid=1] \\ ∆n=-\frac{1}{2} \end{gathered}$$
$$\begin{gathered} q_p-q_v=-\frac{1}{2}\times R\times 300 \\ {q}_p-q_v=-150R \end{gathered}$$
Option (2)



Regards!