Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2(g) + O2(g)
—>2H20(9)

ΔH°r = -483.64 kJ

Explanation:

Let's consider the following balanced equation.

2 H₂(g) + O₂(g)  ⇒ 2 H₂O(g)

We can calculate the standard enthalpy change of the reaction (ΔH°r) using the following expression.

ΔH°r = ∑ΔH°f(p) × np - ∑ΔH°f(r) × nr

where

ΔH°f: standard heat of formation

n: moles

p: products

r: reactants

ΔH°r = ΔH°f(H₂O(g)) × 2 mol - ΔH°f(H₂(g)) × 2 mol - ΔH°f(O₂(g)) × 1 mol

ΔH°r = (-241.82 kJ/mol) × 2 mol - 0 kJ/mol × 2 mol - 0 kJ/mol × 1 mol

ΔH°r = -483.64 kJ

i believe it is c.) 1575 j.