Consider the following reaction occurring in an automobile 2 C 8 H 18( g )+25 O 2( g ) longrightarrow 16 CO 2( g )+18 H 2 O ( g ) the sign of Δ H, Δ S and Δ G would be

Question

Consider the following reaction occurring in an automobile 2 C 8 H 18( g )+25 O 2( g ) longrightarrow 16 CO 2( g )+18 H 2 O ( g ) the sign of Δ H, Δ S and Δ G would be (A) +, -, + (B) -, +, - (C) -, +, + (D) +, +, -

Tardigrade Admin · Accepted Answer

This is combustion reaction, which is always exothermic hence

Δ H = - v e

As the no. of gaseous molecules are increasing hence entropy increases now

Δ G = Δ H - T Δ S

For a spontaneous reaction

Δ G = - v e

Which is possible in this case as

Δ H = - v e

and

Δ S = + v e

Q. Consider the following reaction occurring in an automobile 2C8​H18​(g)+25O2​(g)⟶16CO2​(g)+18H2​O(g) the sign of ΔH,ΔS and ΔG would be

+, -, +

-, +, -

-, +, +

+, +, -

This is combustion reaction, which is always exothermic hence ΔH=−ve As the no. of gaseous molecules are increasing hence entropy increases now ΔG=ΔH−TΔS For a spontaneous reaction ΔG=−ve Which is possible in this case as ΔH=−ve and ΔS=+ve

Q. Consider the following reaction occurring in an automobile $2 C_{8} H_{18} (g) + 25 O_{2} (g) ⟶ 16 C O_{2} (g) + 18 H_{2} O (g)$ the sign of $Δ H, Δ S$ and $Δ G$ would be

This is combustion reaction, which is always exothermic hence
$Δ H = - v e$
As the no. of gaseous molecules are increasing hence entropy increases now $Δ G = Δ H - T Δ S$
For a spontaneous reaction $Δ G = - v e$
Which is possible in this case as $Δ H = - v e$ and $Δ S = + v e$