Use standard enthalpies of formation to calculate ΔH∘rxn for the following reaction______.2H2S(g)+3O2(g)→2H2O(l)+2SO2(g) ΔH∘rxn =

We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.

In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.

The standard enthalpies of the molecules above are as follows:

H2S = -20.63KJ/mol

H2O = -285.8KJ/mol

SO2 = -296.84KJ/mol

O2 = 0KJ/mol

ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3

 ΔfH⦵(O2)]

ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]

-[ 3 × -20.63)]

= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol

fichoh fichoh · Answer 2 · 2021-11-12T06:42:23+01:00

The ΔHrxn for the reaction is the difference between the sum of the standard enthalpies of formation for the product and reactant. Hence. ΔHrxn = - 1124.1 kj/mol

Using following standard enthalpy values :

ΔH°f (H2S) = -20.15 kJ/mol;

ΔH°f (O2) = 0 kJ/, mol;

ΔH°f (H2O) = -285.8 kJ/mol;

ΔH°f (SO2) = -296.4 kJ/mol

The ΔHrxn can be calculated thus :

ΔHf product - ΔHf reactant

Sum of ΔHf of the product :

2(-285.8) + 2(-296.4) = - 1164.4 kj/mol

Sum of ΔHf of the reactant :

2(-20.15) + 3(0) = - 40.30 kj/mol

The ΔHrxn goes thus :

-1164.4 - (-40.30) = -1164.4 + 40.30 = -1124.1 kj/mol

Therefore, the ΔH for the reaction is -1124.1 kj/mol

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