Complete Question
10 g of Compound X with molecular formula [tex]C_4 H_8[/tex] are burned in a constant-pressure calorimeter containing 45g of water at [tex]25^oC[/tex]. The temperature of the water is observed to rise by 2.432. (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 at [tex]25^oC[/tex]
Answer:
The value is [tex]\Delta H^o_f =-2571 \ J /mol[/tex]
Explanation:
From the question we are told that
The mass of compound X is [tex]m = 10 \ g = 0.010 \ kg[/tex]
The mass of water is [tex]m_w = 45 \ g = 0.045 \ kg[/tex]
The temperature of water is [tex]T_w = 25^oC[/tex]
The change in the temperature of water is [tex]\Delta T = 2.432 ^oC[/tex]
Generally the heat adsorbed by water is mathematically represented as
[tex]Q_{w } = c * m _w * \Delta T[/tex]
Here c is the specific heat of water with value [tex]c = 4186 \ J/kg\cdot ^oC[/tex]
[tex]Q_{w } = 4186 * 0.045 * 2.432[/tex]
=> [tex]Q_{w } = 458.1 \ J[/tex]
Given that the total heat that was generated by the reaction is absorbed by water then
[tex]\Delta H _{rxn} = -Q_w[/tex]
The negative sign shows that the heat was absorbed
[tex]\Delta H _{rxn} = -458.1 \ J[/tex]
Generally the number of moles of the compound X available is mathematically represented as
[tex]n = \frac{m}{Z}[/tex]
Here Z is the molar mass of compound X the value is [tex]Z = 56.11 \ g/mol[/tex]
[tex]n = \frac{m}{Z}[/tex]
=> [tex]n = \frac{10}{56.11}[/tex]
=> [tex]n = 0.1782 \ mols[/tex]
Generally the standard heat of formation of Compound X is mathematically represented as
[tex]\Delta H^o_f = \frac{\Delta H_{rxn} }{ n}[/tex]
=> [tex]\Delta H^o_f = \frac{-458.1 }{ 0.1782}[/tex]
=> [tex]\Delta H^o_f = \frac{-458.1 }{ 0.1782}[/tex]
=> [tex]\Delta H^o_f =-2571 \ J /mol[/tex]
