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.
5 answers
Did you give the reaction below?
yeah its here
HNO3(aq) + NaOH(s) -----> NaNO3(aq) + H2O(l)
HNO3(aq) + NaOH(s) -----> NaNO3(aq) + H2O(l)
You gotta be kidding. This has nothing to do with the conversion of Ni to Ni(CO)4 does it?
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.
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.
sorry, I pasted the wrong reaction. thanks for the help
3 answers