Question

An ice calorimeter measures quantities of heat by the quantity of ice melted. How many grams of ice would be melted by the heat released in the complete combustion of 1.60 L of propane gas, measured at and 735 mmHg? [Hint: What is the standard molar enthalpy of combustion of C3H8(g)?]

Answer 1

The question is incomplete, here is the complete question:

An ice calorimeter measures quantities of heat by the quantity of ice melted. How many grams of ice would be melted by the heat released in the complete combustion of 1.60 L of propane gas, measured at 20.0 °C and 735 mmHg? [Hint: What is the standard molar enthalpy of combustion of C₃H₈(g)?]

Answer: The mass of ice that would be melted is 425.52 grams

Explanation:

• To calculate the moles of propane, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735 mmHg

V = Volume of the gas = 1.60 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of propane = ?

Putting values in above equation, we get:

$735mmHg\times 1.60L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735\times 1.60}{62.3637\times 293}=0.064mol$

• The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(C_3H_8(g))})+(5\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(C_3H_8(g))}=-103.8kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-393.5))+(4\times (-285.8))]-[(1\times (-103.8))+(5\times (0))]\\\\\Delta H_{rxn}=-2219.9kJ$

By Stoichiometry of the reaction:

When 1 mole of propane is combusted, the heat released is 2219.9 kJ

So, when 0.064 moles of propane is combusted, the heat released will be = $\frac{2219.9}{1}\times 0.064=142.07kJ$

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

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

where,

$q$ = amount of heat released = 142.07 kJ

n = number of moles of ice = ?

$\Delta H_{fusion}$ = molar heat of fusion = 6.01 kJ/mol

Putting values in above equation, we get:

$6.01kJ/mol=\frac{142.07kJ}{n}\\\\n=\frac{142.07kJ}{6.01kJ/mol}=23.64mol$

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

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

Molar mass of ice = 18 g/mol

Moles of ice = 23.64 moles

Putting values in above equation, we get:

$23.64mol=\frac{\text{Mass of ice}}{18g/mol}\\\\\text{Mass of ice}=(23.64mol\times 18g/mol)=425.52g$

Hence, the mass of ice that would be melted is 425.52 grams