Question

Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH = -196 kJ Calculate the value of q (kJ) in this exothermic reaction when 2.50 g of hydrogen peroxide decomposes at constant pressure.

Answer 1

Answer: The amount of heat released is -7.203 kJ

Explanation:

The given chemical equation follows:

$2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ$

To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:

When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ

So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be $\frac{-196}{2}\times 1=-98kJ$

• To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of hydrogen peroxide = 2.50 g

Molar mass of hydrogen peroxide = 34 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol$

• To calculate the heat of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released

n = number of moles = 0.0735 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -98 kJ/mol

Putting values in above equation, we get:

$-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ$

Hence, the amount of heat released is -7.203 kJ