Question

At constant pressure for which of the reactions shown below should Delta H be greater than Delta E°?
1.2 SO2(g) + O2(g) - 2 SO3(g)
II. C5H12(9) + 8 O2(g) - 5 CO2(g) + 6 H20(1)
III. H2(g) + Br2(g) → 2 HBr(9),
IV. N204(9) - 2 NO2(g)

Answer 1

Answer: $N_2O_4(g)\rightarrow 2NO_2(g)$ has higher value of $\Delta H$ than $\Delta E$

Explanation:

The realtion between enthalpy change and internal energy change is:

$\Delta H=\Delta E+\Delta n_gRT$

where,

=  enthalpy of the reaction

= internal energy of the reaction

$\Delta n_g$ = change in the gaseous moles of the reaction = Moles of product - Moles of reactant

R = gas constant

T = temperature

1. $2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

$\Delta n_g$ = 2-3 = -1

2. $C_5H_{12}(g)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)$

$\Delta n_g$ = 5-9 = -4

3. $H_2(g)+Br_2(g)\rightarrow 2HBr(g)$

$\Delta n_g$ = 2-2 = 0

4. $N_2O_4(g)\rightarrow 2NO_2(g)$

$\Delta n_g$ = 2-1 = 1

Thus reaction . $N_2O_4(g)\rightarrow 2NO_2(g)$ has higher value of $\Delta H$ than $\Delta E$

Answer 2

At constant pressure the given reaction that shows ΔH would be greater than ΔE is;

IV); N₂0₄(g) → 2NO₂(g)

In chemistry, he meaning of ΔH and ΔE are;

- ΔH means the change in enthalpy, which is heat content.

- ΔE means the change in energy of a system.

Now, for a closed system the relationship between ΔH and ΔE is given by;

ΔH = ΔE + PΔV

where PΔV = ΔnRT

Thus; ΔH = ΔE + ΔnRT

where Δn = number of moles of product in the gaseous state - number of moles of reactants.

Thus, if Δn is negative, then ΔE would be greater than ΔH but if Δn is positive, then ΔH would be greater than ΔE

I) 2SO₂(g) + O₂(g) ⇒ 2SO₃(g)

number of moles of product = 2

number of moles of reactants = 2 + 1 = 3

Δn = 2 - 3

Δn = -1 and so ΔE would be greater than ΔH

II) C₅H₁₂(g) + 8O₂(g) ⇒ 5CO₂(g) + 6H₂0(l)

number of moles of product = 5 (6 is not used because water is in liquid and not gas)

number of moles of reactants = 8 + 1 = 9

Δn = 5 - 9

Δn = -4 and so ΔE would be greater than ΔH

III) H₂(g) + Br₂(g) → 2HBr(g)

number of moles of product = 2

number of moles of reactants = 1 + 1 = 2

Δn = 2 - 2

Δn = 0 and so ΔE would be equal to ΔH

IV) N₂0₄(g) → 2NO₂(g)

number of moles of product = 2

number of moles of reactants = 1

Δn = 2 - 1

Δn = +1 and so ΔH would be greater than ΔE

Read more at; https://brainly.com/question/15701463