Question

Consider the reaction described by the following chemical equation.

2 HN3(1) + 2 NO(g) — H2O2(1) + 4 N, (g)
What is the enthalpy change associated with the production of 1 mol of H, O, if a reaction that produces
2.50 g of H, O, releases 65.9 kJ of heat?

Answer 1

Q = -897 kJ/mol

Step-by-step explanation:

From the given information:

The heat released Q = -65.9 kJ

To start with the molar mass of
`H_2O_2` = 2 × (molar mass of H) + 2 × (molar mass of O)

= (2 × 1.008) + (2 × 16.0 )

= 34.016 g/mol

However, given that:

mass of
`H_2O_2` 2.50 g

The number of moles of
`H_2O_2` =
`(mass)/(molar \ mass)`

`= (2.5)/(34.016)`

`= 7.349 * 10^(-2) \ mol`

Finally; Using the formula:

`\Delta H = (Q)/(number \ of \ moles)\\ \\ Q = (-65.9 \ kJ)/(7.349 * 10^(-2) \ mol)`

Q = -897 kJ/mol