Question

The standard Gibbs energy change for the formation of propane $\left(C_3{H}_{8\left(g\right)}\right)$ at 298 K is

$[{\Delta }_f{H}^{\theta }$ for propane $=-103.85$ kJ mole${}^{-1},$ $S_m^{\theta }{C}_3{H}_{8\left(g\right)}=270.2$ JK${}^{-1}$mol${}^{-1}$, $S_m^{\theta }{H}_{2\left(g\right)}=130.68$ JK${}^{-1}$mol${}^{-1}$, $S_m^{\theta }{C}_{\left(graphite\right)}=5.742$ JK${}^{-1}$mol${}^{-1}]$

• A. $-23.4$ kJ
• B. $-44.4$ kJ
• C. $-54.4$ kJ
• D. $-104.5$ kJ

Answer 1

The correct option is A $-23.4$ kJ

Formation of propane
$3C+4H_2\to C_3{H}_8$
$\Delta S_f$ for propane $=270.2-4\times 130.68-3\times 5.742=-269.746$ JK${}^{-1}$mol${}^{-1}$
$\Delta H=-103.85$ kJK${}^{-1}$ mol ${}^{-1}$

$\Delta G=\Delta H-T\Delta S$
$\Delta G=-103.85-298\times (-269.746)/1000=-23.4$kJ