Question

Sulfur reacts with oxygen to form sulfur dioxide (SO2(g), Delta. Hf = â€"296. 8 kJ/mol) according to the equation below. Upper S (s) plus upper O subscript 2 (g) right arrow upper S upper o subscript 2 (g). What is the enthalpy change for the reaction? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants. â€"593. 6 kJ â€"296. 8 kJ 296. 8 kJ 593. 6 kJ.

Answer 1

The enthalpy of this reaction is -296. 8 kJ/mol.

The equation of the reaction is;

S(s) + O2(g) -----> SO2(g)

Recall that the enthalpy of the reaction can be obtained from the formula;

ΔHrxn = ΔHf(products) - ΔHf(reactants)

We know that;

ΔHf S(s) = 0 KJ/mol

ΔHf O2(g) =  0 KJ/mol

ΔHf SO2(g) = -296. 8 kJ/mol

Hence;

ΔHrxn =  -296. 8 kJ/mol - [0 KJ/mol + 0 KJ/mol]

ΔHrxn = -296. 8 kJ/mol

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