Question

A 1.764-g sample of heptanoic acid, C7H14O2 (130.19 g/mol) was burned in a bomb calorimeter with excess oxygen. The temperature of the calorimeter and the water before combustion was 23.68 °C; after combustion the calorimeter and the water had a temperature of 32.12 °C. The calorimeter had a heat capacity of 500 J/K, and contained 1.462 kg of water. Use these data to calculate the molar heat of combustion (in kJ) of heptanoic acid.

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in....typed format....

$\Rightarrow$Answer:

Molar heat of combustion of heptanoic acid ( C₇H₁₄O₂ ) : Q =   4138.52 kJ /mol  ...

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above Answer...

• Given:

1. ​​​​​​​Mass of heptanoic acid : ( C₇H₁₄O₂ ) : m₁ = 1.764 g ( grams )
2. Molar mass of heptanoic acid : ( C₇H₁₄O₂ ) : n₁ = 130.19 g/mol ( grams per mole )
3. Temperature of the calorimeter & water: before combustion (initially ): T_i =   23.68 ^oC =  ( 273.15 + 23.68 ) = 296.83 K ( Kelvin )
4. Final temperature of the calorimeter and water: T_f = 32.12 ^oC = ( 273.15 + 32.12 ) = 305.27 K ( kelvin)
5. Mass of water in the calorimeter: m_h2o = 1.462 kg ( kilo-grams )
6. Heat capacity of the calorimeter: C_cal = 500.0 J / K = 0.500 kJ / K ( kilo-joule per kelvin )
7. We know: Specific heat of Water: C_h2o = 4.186 kJ / kg. K

• ​​​​​​​Step - 1:

$\Rightarrow$ Number of moles of  heptanoic acid : ( C₇H₁₄O₂ ) : =  m₁ / n₁ =  ( 1.764 g ) / ( 130.19 g/mol ) = 0.0135 mol

• Step - 2:

​​​​​​​$\Rightarrow$ Change in temperature of the calorimeter and water : $\Delta$T =  T_f - T_i = ( 305.27 - 296.83 ) = 8.44 K ( Kelvin )

$\Rightarrow$ Heat gained by calorimeter: Q_cal = C_cal x  $\Delta$T =  ( 0.500 kJ/K ) x ( 8.44 K ) = 4.22 kJ

$\Rightarrow$ Heat gained by water, in caloriemter: Q_h2o = m_h2o x C_h2o x  $\Delta$T

= ( 1.462 kg ) x (4.186 kJ / kg. K ) x ( 8.44 K )

= 51.65 kJ ( kilo-joule )

• Step - 3:

Therefore: total heat of combustion: ( Q_comb ) will be equal to the following:

$\Rightarrow$    Q_comb =  Q_cal + Q_h2o =  ( 4.22 + 51.65 ) kJ = 55.87 kJ   ( kilo-joule )

• Step - 4:

​​​​​​​As we can see above, the following : i.e.

$\Rightarrow$ 0.0135 mol ( moles ) of heptanoic acid : ( C₇H₁₄O₂ ) , after combustion gives: 55.87 kJ ( kilo-joule ) of Heat ...

$\Rightarrow$1.0 mol ( moles ) of heptanoic acid : ( C₇H₁₄O₂ ), would give: ( 55.87 kJ ) / ( 0.0135 mol ) = 4138.52 kJ ( approx. ) of heat ...

Therefore:

$\Rightarrow$Molar heat of combustion of heptanoic acid : ( C₇H₁₄O₂ ) =  4138.52 kJ /mol