Asked by Joshua Chang

Calculate the amount of work done for the conversion of 1.00 mole of Ni to Ni(CO)4 in the reaction below, at 75oC. Assume that the gases are ideal. The value of R is 8.31 J/molK.

I know the answer is 8.68*10^3 but I don't know why that is.

Answers

Answered by DrBob222
Did you give the reaction below?
Answered by Joshua Chang
yeah its here

HNO3(aq) + NaOH(s) -----> NaNO3(aq) + H2O(l)
Answered by DrBob222
You gotta be kidding. This has nothing to do with the conversion of Ni to Ni(CO)4 does it?
Answered by DrBob222
Ni + 4CO ==> Ni(CO)4
1 mol + 4 mols = 1 mol
work = -pdV. I assume the pressure is 1 atm
so delta V is -3 since 4 mols is changed to 1 mol.
dV = dnRT/P = -3*0.082*(273+75)/1
I know I used P in atm and R in L-atm/C but I convert at the end.
dV = -85.5 L*atm so -pdV is +85.5
85.5 x 101.3 = 8.66E3 J. Use T and R a little more accurately and probably you will get the 8.68E3 J = work.

Answered by Joshua Chang
sorry, I pasted the wrong reaction. thanks for the help
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