Answer:
[tex]\Delta H_{f,C_3H_4}=276.8kJ/mol[/tex]
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
[tex]\Delta H_{rxn} =- m_wC_w\Delta T[/tex]
We plug in the mass of water, temperature change and specific heat to obtain:
[tex]\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ[/tex]
Now, this enthalpy of reaction corresponds to the combustion of propyne:
[tex]C_3H_4+4O_2\rightarrow 3CO_2+2H_2O[/tex]
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
[tex]\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}[/tex]
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
[tex]\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol[/tex]
Now, we solve for the enthalpy of formation of C3H4 as shown below:
[tex]\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}[/tex]
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
[tex]\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol[/tex]
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