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Write the balanced chemical equation for the enthalpy of formation of nickel (II)...
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- We need to turn both reactants into monatomic gases so that they can react completely
- We need to ionize both reactants with the gain or loss of two electrons
- We need to account for the loss of electromagnetic potential when the charged ions bond and produce a solid
High School Teacher
The enthalpy of formation describes the change in energy that a system of reactants undergoes in order to form the product. In this case, nickel and oxygen are the reactants which undergo a thermodynamic and chemical change, to produce nickel (ii) oxide. Nickel is normally present as a solid, and oxygen as a gas.` Ni(s) + O_2 (g)-> NiO(s)`
There isn't really a generalized way of knowing what the state of the product will be, but a little research shows that nickel(ii) oxide will be a solid. Nickel itself is normally present as a solid, and oxygen as a gas.
In order to solve this problem, you will need several charts. Unfortunately not all of the charts that I have, or was able to find on the internet, gave consistent values; I would recommend that you use only the charts provided in your textbook or by your instructor.
There are several steps that we need to account for;
First, we only need one oxygen atom for every nickel atom; however oxygen is only available in O2 form. Thus we can either double some of our coefficients for nickel, or halve our coefficients for oxygen gas. I have chosen to halve the oxygen gas coefficients.
Enthalpy of atomization for nickel: 431kj
Enthalpy of atomization for O2: 294(.5) = 147kj
Second, we need to take away two electrons from the nickel, and add them to the oxygen.
First and second ionization energies of nickel atoms: 737kj, 1753kj
First and second electron affinities for oxygen: -142kj, 844kj
Third, we account for the lattice energy of the newly bonded NiO: I found a variety of numbers for this value, and have chosen to use -4000kj. Again, I strongly recommend that you use only the data provided in your own textbook for these values.
Thus, 431+147+737+1753-142+844-4000 = -230kj.
Posted by caledon on January 13, 2014 at 3:00 AM (Answer #1)
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