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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

a polar bond is formed when electrons are shared unevenly between atoms