Question

6.000 g of Compound X with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 25.00 kg of water at 25 °C. Thetemperature of the water is observed to rise by 2.656 °C. (You may assume all the heat released by the reaction is absorbed by the water, andnone 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 3 significant digits.100.0XS?oloR23ER

Answer 1

The standard heat of formation of Compound X at 25 °C is approximately
`\(1758.34 \, \text{kJ/mol}\)` (rounded to three significant digits).

To calculate the standard heat of formation
`(\(\Delta H_f^\circ\))` for Compound X, we can use the equation:

`\[ q = mc\Delta T \]`

where:

-
`\( q \)` is the heat absorbed by the water,

-
`\( m \)` is the mass of water,

-
`\( c \)` is the specific heat of water, and

-
`\( \Delta T \)` is the change in temperature.

The heat absorbed by the water is equal to the heat released by the combustion of Compound X:

`\[ q = \Delta H_f^\circ * n \]`

where:

-
`\( \Delta H_f^\circ \)` is the standard heat of formation per mole of Compound X, and

-
`\( n \)` is the number of moles of Compound X.

First, calculate the number of moles of Compound X:

`\[ \text{Molar mass of }C_3H_4 = (3 * \text{atomic mass of C}) + (4 * \text{atomic mass of H}) \]`

Now, determine the number of moles
`(\( n \))`:

`\[ n = \frac{\text{mass of Compound X}}{\text{molar mass of Compound X}} \]`

Next, calculate the heat
`(\( q \))` absorbed by the water using the first equation:

`\[ q = mc\Delta T \]`

Finally, use the second equation to find
`\( \Delta H_f^\circ \)`:

`\[ \Delta H_f^\circ = (q)/(n) \]`

Now, let's perform the calculations:

`\[ \text{Molar mass of } C_3H_4 = (3 * 12.01 \, \text{g/mol}) + (4 * 1.01 \, \text{g/mol}) \]`

`\[ = 36.03 + 4.04 = 40.07 \, \text{g/mol} \]`

`\[ n = \frac{6.000 \, \text{g}}{40.07 \, \text{g/mol}} \]`

`\[ \approx 0.1497 \, \text{mol} \]`

`\[ q = (25.00 \, \text{kg}) * (4.18 \, \text{J/g}^\circ\text{C}) * (2.656 \text{\textdegree C}) \]`

`\[ = 263.124 \, \text{kJ} \]`

`\[ \Delta H_f^\circ = \frac{263.124 \, \text{kJ}}{0.1497 \, \text{mol}} \]`

`\[ \approx 1758.34 \, \text{kJ/mol} \]`

So, the standard heat of formation of Compound X at 25 °C is approximately
`\(1758.34 \, \text{kJ/mol}\)` (rounded to three significant digits).

Answer 2

First we need to determine the heat energy absorbed by the water:

`\begin{gathered} q_(H2O)=mc\Delta T \\ q:heat\text{ }energy\text{ }absorbed(J) \\ m:mass(kg)=25.00\text{ }kg \\ c:specific\text{ }heat\text{ }capacity(4184Jkg^(-1)K^(-1)) \\ \Delta T:temperature\text{ }in\text{ }K\text{ }(273.15K+2.656\degree C=275.806K) \end{gathered}`

We will determine the heat energy absorbed by the water by substituting the given values into the equation:

`\begin{gathered} q_(H2O)=25.00kg*4184Jkg^(-1)K^(-1)*275.806K \\ q_(H2O)=28,849,307.6\text{ }J \end{gathered}`

The energy released by the reaction is equivalent to the energy absorbed by the water:

`q_(rxn)=-q_(H2O)`

Therefore the amount of energy released by the reaction is -28,849,307.6J. We will convert this energy to kJ so we divide by 1000. Energy released is -28,849.31 kJ

We need to convert the mass of the compound to moles of the compound:

`\begin{gathered} _nC_3H_4=\frac{mass}{molar\text{ }mass} \\ _nC_3H_4=(6.0g)/(40.06gmol^(-1)) \\ _nC_3H_4=0.15mol \end{gathered}`

The enthalpy of formation for the reaction is:

`\begin{gathered} \Delta H_(rxn)=(q_(rxn))/(n) \\ \Delta H_(rxn)=(-28,849.31kJ)/(0.15mol) \\ \Delta H_(rxn)=-192,328kJmol^(-1) \\ \Delta H_(rxn)=-192,000\text{ }kJmol^(-1) \end{gathered}`

Answer: The standard heat of formation or enthalpy of formation of compound X is -192,000kJ/mol.