Question

A mixture of 2 moles of carbon monoxide and one
mole of oxygen in a closed vessel is ignited to get
carbon dioxide. If AH is the enthalpy change and AE
is the change in internal energy, then
a) AH> AE
b) AH<AE
c) AHEAE
d) not definite​

Answer 1

Answer:

b. ∆H<∆E

Explanation:

2CO(g)+O2(g)------> 2CO2(g)

∆n(gas)= 2-3 = -1

∆H= ∆E+∆n(gas)RT

∆H= ∆E -RT

hence ∆H<∆E

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Answer 2

If ΔH is the enthalpy change and  ΔE  is the change in internal energy, then $\Delta H$< $\Delta E$

Explanation:

The balanced chemical reaction is:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

$\Delta H=\Delta E+\Delta n_g\times RT$

where,

$\Delta H$ =  enthalpy of the reaction

$\Delta E$= internal energy of the reaction

$\Delta n_g$ = change in the moles of the reaction = Moles of product - Moles of reactant = 2 - 3 = -1

R = gas constant

T = temperature

Putting in the values we get:

$\Delta H=\Delta E+(-1)\times RT$

$\Delta H=\Delta E-\times RT$

Thus $\Delta H$< $\Delta E$

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