Answer to Question #100292 in General Chemistry for Beverlie

Question

1. In an experiment, 9.80 g of olive oil at 80.0C are quickly transferred to water in a calorimeter at 18.0°C. The heat capacity of the water is 4.18 J/gC. The final equilibrium temperature is 22.8°C. What is the approximate specific heat capacity of the olive oil? 5. ) Given the standard enthalpies of formation for the following substances, determine the change in enthalpy for the combustion of 1.0 mol of propane, C3H8. Substance ΔH°f in kJ/mol C3H8(g) 103.8 CO2(g) 393.5 H2O(l) 285.8 6. Calculate the energy released as heat when 45.8 g of Fe(s) is combined with 27.9g of O2(g) if the ΔH°rxn is -1648 kJ/mol. The balanced equation for the reaction is 4 Fe(s) + 3O2(g) = 2 Fe2O3(s)

Answer 1

1. Heat gained = Heat lost

$m_{water}s_{water}\Delta T = m_{olive}s_{olive}\Delta T$

$m*4.18*4.8=9.8*s_{oil}*(80-22.8)$

$\implies s_{oil}=m*2.047/57.2$

$=m*0.0358J/g^°C$

where m is mass (in grams) of water in calorimeter.

5. Combustion of 1 mol propane takes place according to the following chemical reaction:

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

Thus change in enthalpy for this reaction is:

$\Delta H= H_{f(products)}-H_{f(reactants)}$

$=[3H_{f(CO_2​)​}+4H_{f(H_2​O)}​]−$$[H_{f(C_3​H_8​)​}+5H_{f(O_2​)}​]$

$= 3*(-393.5)+ 4*(-285.8)-(-103.8)-5*0$

$\implies \Delta H= -2219.9 kJ$

Thus, heat released in this reaction is $2219.9 kJ$ (Answer)

6. Given $\Delta H_{rxn} = -1648 kJ$

for 4 moles of $Fe(s)$

Thus heat released on combustion of 1 mol or 55.8g $Fe(s)$ will be $1648/4=412kJ$

Thus, heat released on combustion of 45.8g $Fe(s)$ is $412*45.8/55.8=35.44kJ$ (Answer)