q = mass water x specific heat water x (Tfinal-Tinitial).
Then q/0.025 mol = enthalpy/mol.
You will get better responses if you limit your questions to one per post AND if you show some work.
A sample consisting of 0.025 mol of a solid compound is placed inside a metal cylinder. The cylinder was immersed in a water bath consisting of 10.00 kg of water at an original temperature of 25.00 °C. When the compound decomposes, the temperature of the water rises to 29.52 °C. Assume that the heat capacity of the cylinder is negligible. The specific heat of water is 4.184 J/(°C.g).
a. Calculate the heat evolved (in kJ) during the decomposition reaction, and from that determine the molar enthalpy change of the decomposition reaction.
b. The decomposition reaction may be written in this form:
C12O36H20N12 (s) → 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O (l)
In the enthalpy diagram below, the arrows represent standard enthalpy reactions. (3) denotes the standard enthalpy of decomposition of C12O36H20N12 (s). Provide an appropriate label for the arrows (1), (2) and (4).
(3) C12O36H20N12 (s) --> 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O(l)
(1) 12C (s) + 18O2 (g) + 6N2 (g) + 10H2 (g) --> 12CO2 (g) + 6O2 (g) + 6N2 (g) + 10H2 (g)
(2) 12CO2 (g) + 6O2 (g) + 6N2 (g) + 10H2 (g) --> 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O(l)
(4) 12C (s) + 18O2 (g) + 6N2 (g) + 10H2 (g) --> C12O36H20N12 (s)
c. Use the enthalpy diagram provided above and apply Hess’s Law to determine the standard enthalpy of formation for C12O36H20N12 (s) using the results from part (a) and the following values:
The standard enthalpy of formation of gaseous carbon dioxide is -393.5 kJ/mol
The standard enthalpy of formation of liquid water is -286 kJ/mol
(Strictly speaking, standard enthalpy is defined for a reaction that occurs at 25 °C, but for the purpose of this question you may ignore the difference in temperature.)
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