Question

In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.369 x 10^3 g of water. During the combustion the temperature increases from 22.95 to 26.05°C. The heat capacity of water is 4.184 J g-1°C-1. The heat capacity of the calorimeter was determined to be 916.9 J/°C.
1. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of diphenylacetylene based on these data.

Answer 1

Answer:

the Molar heat of  Combustion  of  diphenylacetylene $(C_{14}H_{10})$  = $-6.931 *10^3 \ kJ/mol$

Explanation:

Given that:

mass of diphenylacetylene $(C_{14}H_{10})$ = 0.5297 g

Molar Mass of diphenylacetylene $(C_{14}H_{10})$ = 178.21 g/mol

Then number of moles of diphenylacetylene $(C_{14}H_{10})$  = $\frac{mass}{molar \ mass}$

= $\frac{0.5297 \ g }{178.24 \ g/mol}$

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$  = Heat absorbed by $H_2O$ + Heat absorbed  by the calorimeter

Heat liberated  by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$  =  msΔT + cΔT

= 1369 g  × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$  = 20598.87 J

Heat liberated by 1 mole of  diphenylacetylene $(C_{14}H_{10})$ will be = $\frac{20598.87 \ J}{0.002972 \ mol}$

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of  Combustion  of  diphenylacetylene $(C_{14}H_{10})$  = $-6.931 *10^3 \ kJ/mol$

Answer 2

Answer:

Explanation:

To solve this type of problem, we have to keep in mind that the calorimeter will absorb some part of the heat released during combuston, and that the water in the bomb calorimeter will absorb the rest of the heat, assuming no heat is lost to the surroundings.

Thus to calculate the heat of combustion of diphenylacetylene in this question we will compute these 2 heats and add them together. Since we are asked to calculate the molar heat we will divide  this total heat by the mol of sample.

q water = m x c x ΔT

where m is the water mass, c the specific heat capacity of water and ΔT the change in temperature.

q water = 1.369 x 10³ g  x 4.184 J/gºC x ( 26.05- 22.95 ) ºC

             = 1.78 x 10⁴ J

q calorimeter = C x  ΔT

where C is the calorimeter specific heat, and ΔT is the change in temperature.

q calorimeter = 916.9 J/ºC x  ( 26.05- 22.95 )ºC = 2.84 x 10³ J

q total = 1.78 x 10⁴ J  +  2.84 x 10³ J = 2.06 x 10⁴ J

mol sample = mass / MW

molar mass C₁₄H₁₀ = 178.23 g/mol

mol C₁₄H₁₀  = 0.5297 g / 178.23 g /mol = 2.97 x 10⁻³ mol

molar heat of combustion C₁₄H₁₀ =   2.06 x 10⁴ J /  2.97 x 10⁻³ mol

                                                      =    6.93 x 10⁶ J / mol = 6.93 x 10³ kJ/mol