Question

The combustion of 1.685 g of propanol (C3H7OH) increases the temperature of a bomb calorimeter from 298.00 K to 302.16 K. The heat capacity of the bomb calorimeter is 13.60 kJ/K . Determine ΔH for the combustion of propanol to carbon dioxide gas and liquid water. g

Answer 1

Answer:

ΔH =  - 2020.57 kJ/mol

Explanation:

Given that :

mass of propanol = 1.685 g

the molar molar mass = 60 g/mol

Thus; the number of  moles = mass/molar mass

= 1.685 g/60 g/mol

= 0.028 g/mol

However ;

ΔH = heat capacity C × Δ T

Given that:

The temperature increases from  298.00 K to 302.16 K.

Then ;

Δ T = 302.16 K - 298.00 K

Δ T = 4.16 K

heat capacity C = 13.60 kJ/K

∴

ΔH = 13.60 kJ/K × 4.16 K

ΔH =  56.576 kJ

The equation of the given reaction can be represented as :

$C_3H_7OH_{(l)}+\dfrac{3}{2}O_{2(g)} \to 3CO_{2(g)} +4H_2O_{(l)}$

Thus for 0.028 mol of heat liberated; ΔH =  56.576 kJ

For 1 mole of heat liberated now:

ΔH =  56.576 kJ/0.028 mol

ΔH =  2020.57 kJ/mol

SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;

ΔH =  - 2020.57 kJ/mol

Answer 2

Answer:

Explanation:

,