A mixture of 2 mol of carbon monoxide and 1 mol of oxygen in a closed vessel is ignited to get carbon dioxide, then

Question

A mixture of 2 mol of carbon monoxide and 1 mol of oxygen in a closed vessel is ignited to get carbon dioxide, then (A) Δ H>Δ E (B) Δ H < Δ E (C) Δ H=Δ E (D) not definite

Tardigrade Admin · Accepted Answer

2CO(.g.)+O2(.g.) arrow 2CO2(.g.) Δ ng=2-(.2+1.)=-1 Δ H=Δ E+Δ ngRT or Δ H=Δ E-1.RT i.e., Δ H < Δ E

Q. A mixture of 2 mol of carbon monoxide and 1 mol of oxygen in a closed vessel is ignited to get carbon dioxide, then

$Δ H > Δ E$

$Δ H < Δ E$

$Δ H = Δ E$

not definite

$2 CO (g) + O_{2} (g) \to 2 C O_{2} (g)$

$Δ n_{g} = 2 - (2 + 1) = - 1$

$Δ H = Δ E + Δ n_{g} RT$

or $Δ H = Δ E - 1. RT$

i.e., $Δ H < Δ E$

Q. A mixture of 2 mol of carbon monoxide and 1 mol of oxygen in a closed vessel is ignited to get carbon dioxide, then

ΔH>ΔE

ΔH<ΔE

ΔH=ΔE

not definite

2CO(g)+O2​(g)→2CO2​(g) Δng​=2−(2+1)=−1 ΔH=ΔE+Δng​RT or ΔH=ΔE−1.RT i.e., ΔH<ΔE

$Δ H > Δ E$

$Δ H < Δ E$

$Δ H = Δ E$

$2 CO (g) + O_{2} (g) \to 2 C O_{2} (g)$

$Δ n_{g} = 2 - (2 + 1) = - 1$

$Δ H = Δ E + Δ n_{g} RT$

or $Δ H = Δ E - 1. RT$

i.e., $Δ H < Δ E$