For the reaction, C 3 H 8( g )+5 O 2( g ) longrightarrow 3 CO 2( g )+4 H 2 O ( l ) at constant temperature, Δ H-Δ E is

Question

For the reaction, C 3 H 8( g )+5 O 2( g ) longrightarrow 3 CO 2( g )+4 H 2 O ( l ) at constant temperature, Δ H-Δ E is (A) RT (B) - 3RT (C) 3RT (D) - RT

Tardigrade Admin · Accepted Answer

C 3 H 8( g )+5 O 2( g ) longrightarrow 3 CO 2( g )+4 H 2 O ( l ) Δ ng=np-nr =3-6=-3 Δ H=Δ E+Δ ng R T Δ H=Δ E-3 R T Δ H-Δ E=-3 R T

Q. For the reaction,
$C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (l)$
at constant temperature, $Δ H - Δ E$ is

RT

- 3RT

3RT

- RT

$C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (l)$
$Δ n_{g} = n_{p} - n_{r}$
$= 3 - 6 = - 3$
$Δ H = Δ E + Δ n_{g} RT$
$Δ H = Δ E - 3 RT$
$Δ H - Δ E = - 3 RT$

Q. For the reaction, C3​H8​(g)+5O2​(g)⟶3CO2​(g)+4H2​O(l) at constant temperature, ΔH−ΔE is

RT

- 3RT

3RT

- RT

C3​H8​(g)+5O2​(g)⟶3CO2​(g)+4H2​O(l) Δng​=np​−nr​ =3−6=−3 ΔH=ΔE+Δng​RT ΔH=ΔE−3RT ΔH−ΔE=−3RT

Q. For the reaction,
$C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (l)$
at constant temperature, $Δ H - Δ E$ is

$C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (l)$
$Δ n_{g} = n_{p} - n_{r}$
$= 3 - 6 = - 3$
$Δ H = Δ E + Δ n_{g} RT$
$Δ H = Δ E - 3 RT$
$Δ H - Δ E = - 3 RT$