Question
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.
I know the answer is 8.68*10^3 but I don't know why that is.
Answers
DrBob222
Did you give the reaction below?
Joshua Chang
yeah its here
HNO3(aq) + NaOH(s) -----> NaNO3(aq) + H2O(l)
HNO3(aq) + NaOH(s) -----> NaNO3(aq) + H2O(l)
DrBob222
You gotta be kidding. This has nothing to do with the conversion of Ni to Ni(CO)4 does it?
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.
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.
Joshua Chang
sorry, I pasted the wrong reaction. thanks for the help