A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide.If Δ H is the enthalpy change and Δ E is the change in internal energy, then,

Question

A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide.If Δ H is the enthalpy change and Δ E is the change in internal energy, then, (A) Δ H > Δ E (B) Δ H < Δ E (C) Δ H = Δ E (D) the relationship depends on the capacity of the vessel

Tardigrade Admin · Accepted Answer

2 CO + O 2 longrightarrow 2 CO 2 Δ H=Δ E+Δ n R T Δ n= No. of moles of gaseous product - No. of moles of gaseous reactant ∴ Δ n=2-3=-1 Δ H=Δ E-R T ∴ Δ H< Δ E

Q. A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide.If $Δ H$ is the enthalpy change and $Δ E$ is the change in internal energy, then,

$Δ H > Δ E$

$Δ H < Δ E$

$Δ H = Δ E$

the relationship depends on the capacity of the vessel

$2 CO + O_{2} ⟶ 2 C O_{2}$

$Δ H = Δ E + Δ n RT$

$Δ n =$ No. of moles of gaseous product - No. of moles of gaseous reactant

$∴ Δ n = 2 - 3 = - 1$

$Δ H = Δ E - RT$

$∴ Δ H < Δ E$

Q. A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide.If ΔH is the enthalpy change and ΔE is the change in internal energy, then,

ΔH>ΔE

ΔH<ΔE

ΔH=ΔE