For the following reactions is delta Hrnx equal to delta Hf of the products :
a. CaCO3(g)----> CaO + CO2(g)
b. 2Na(s)+ Cl2(g) ---> 2NaCl(s)
c. C(s)+ O2(g) -----> CO2(g)
d. CO(g)+ 1/2O2(g)---> CO2(g)
e. Na(s)+ 1/2Cl2(l) ---> NaCl(s)
f. Na(s)+ 1/2Cl2(s) ---> NaCl(s)
i Know that A and D are not since they are compounds, I cant seem to get the answer, because i cant really distinguish between the rest.
By the definition of heat of formation (when one mole of compound is formed from the elements at 25C), the only possible equations is c. For f) I have never seen chlorine as a solid.
oops its Cl (g)
The f is from the elements, yielding one mole.
To determine whether delta Hrnx (the change in enthalpy of a reaction) is equal to delta Hf (the standard enthalpy of formation) of the products, you need to compare the stoichiometric coefficients of the reactants and products in the balanced equation.
Delta Hf represents the change in enthalpy that occurs when one mole of a compound is formed from its constituent elements in their standard states at a given temperature and pressure. It is typically given as a standard value in a reference table.
Let's analyze each reaction:
a. CaCO3(g) --> CaO(s) + CO2(g)
In this reaction, CaCO3 decomposes into CaO and CO2. The enthalpy change for this reaction, delta Hrnx, could be measured experimentally. However, delta Hf of the products (CaO and CO2) cannot be directly determined from this given reaction. So, the answer is not clear based on the given information.
b. 2Na(s) + Cl2(g) --> 2NaCl(s)
In this reaction, 2 moles of Na and 1 mole of Cl2 react to form 2 moles of NaCl. Since the stoichiometric coefficients match the formation equation for NaCl (2Na + Cl2 --> 2NaCl), we can conclude that delta Hrnx will be approximately equal to 2 times the delta Hf value of NaCl.
c. C(s) + O2(g) --> CO2(g)
In this reaction, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. The stoichiometric coefficients are consistent with the formation equation for CO2 (C + O2 --> CO2), so delta Hrnx will be approximately equal to the delta Hf value of CO2.
d. CO(g) + 1/2O2(g) --> CO2(g)
Similarly to reaction c, the stoichiometric coefficients in this equation match the formation equation for CO2 (CO + 1/2O2 --> CO2). Therefore, delta Hrnx will be approximately equal to the delta Hf value of CO2.
e. Na(s) + 1/2Cl2(l) --> NaCl(s)
In this reaction, 1 mole of Na reacts with 1/2 mole of Cl2 to form 1 mole of NaCl. The stoichiometric coefficients do not directly match the formation equation for NaCl (2Na + Cl2 --> 2NaCl), so delta Hrnx will not be equal to the delta Hf value of NaCl.
f. Na(s) + 1/2Cl2(s) --> NaCl(s)
This reaction is the same as reaction e, but with chlorine in the solid state instead of liquid. Since the stoichiometric coefficients still do not directly match the formation equation for NaCl, delta Hrnx will not be equal to the delta Hf value of NaCl.
In summary, based on the given information, the reactions b, c, and d have delta Hrnx approximately equal to the delta Hf values of their respective products. The reactions a, e, and f do not have a direct relationship between delta Hrnx and delta Hf of the products.