Question

7.00g of Compound X with molecular formula C5H10 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25°C. The temperature of the water is observed to rise by 2.113°C. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound X at 25°C. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits. [Could you please show a step by step way to solve this]

Answer 1

Answer:

The standard heat of formation of Compound X at 25°C is -3095.75 kJ/mol.

Explanation:

Mass of compound X = 7.00 g

Moles of compound X = $\frac{7.00 g}{70 g/mol}=0.100 mol$

Mass of water in calorimeter ,m= 35.00 kg = 35000 g

Change in temperature of the water in calorimeter = ΔT

ΔT = 2.113°C

Specific heat capacity of water ,c= 4.186 J/g °C

Q =  m × c × ΔT

$Q=35000 g\times 4.186 J/g ^oC\times 2.113^oC=309,575.6 J=309.575 kJ$

Heat gained by 35 kg of water is equal to the heat released on burning of 0.100 moles of compound X.

Heat of formation of Compound X at 25°C:

$\frac{-Q}{\text{moles of compound X}}=\frac{-309.575 }{0.100 mol}$

= -3095.75 kJ/mol