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.
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.
You gotta be kidding. This has nothing to do with the conversion of Ni to Ni(CO)4 does it?
Well, I must say, that's quite a "ni-ce" question you've got there! Let's break it down step by step.
First off, we need to determine the change in Gibbs free energy (ΔG) for the reaction. The formula for ΔG is ΔG = ΔH - TΔS, where ΔH is the change in enthalpy and ΔS is the change in entropy. In this case, we'll assume that both ΔH and ΔS are constant with respect to temperature.
Next, we need to calculate ΔH, which is the heat absorbed or released in the reaction. In this case, we have the conversion of 1 mole of Ni to Ni(CO)4, and we need to consider the enthalpy changes associated with each individual reaction step.
The reaction can be divided into two steps:
1. The dissociation of CO gas: CO(g) -> C(s) + O(g)
2. The formation of Ni(CO)4: Ni(s) + 4CO(g) -> Ni(CO)4(g)
For each step, we can look up the standard enthalpy of formation (ΔHf) values for the reactants and products. We find that the ΔHf for Ni(s) is 0 kJ/mol, the ΔHf for CO(g) is -110.5 kJ/mol, and the ΔHf for Ni(CO)4(g) is -566 kJ/mol.
So, the enthalpy change for step 1 is ΔH1 = ΔHf(C(s)) + ΔHf(O(g)) - ΔHf(CO(g)) = 0 kJ/mol + 0 kJ/mol - (-110.5 kJ/mol) = 110.5 kJ/mol.
And the enthalpy change for step 2 is ΔH2 = ΔHf(Ni(CO)4(g)) - [ΔHf(Ni(s)) + 4ΔHf(CO(g))] = -566 kJ/mol - (0 kJ/mol + 4(-110.5 kJ/mol)) = -566 kJ/mol + 442 kJ/mol = -124 kJ/mol.
Now, we can calculate the total enthalpy change for the reaction (ΔHtotal) by summing the enthalpy changes for both steps: ΔHtotal = ΔH1 + ΔH2 = 110.5 kJ/mol + (-124 kJ/mol) = -13.5 kJ/mol.
Finally, we can calculate the work done (w) using the formula w = -ΔHtotal, because the enthalpy change is negative for an exothermic reaction. Thus, w = -(-13.5 kJ/mol) = 13.5 kJ/mol.
Now, to convert the work done to joules, we use the conversion factor 1 kJ = 1000 J. So, the work done is 13.5 kJ/mol * 1000 J/kJ = 13,500 J/mol.
Since we are given that there is 1.00 mole of Ni, the amount of work done for the conversion of 1.00 mole of Ni to Ni(CO)4 is simply 13,500 J/mol * 1 mol = 13,500 J.
And there you have it! The amount of work done for the conversion of 1.00 mole of Ni to Ni(CO)4 is 13,500 J. Now, go out there and "Ni-ce" the day!
To calculate the amount of work done for the conversion of 1.00 mole of Ni to Ni(CO)4, we need to use the ideal gas equation and consider the change in volume during the reaction.
The ideal gas equation is given as:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = gas constant
T = temperature in Kelvin
In this case, we know the value of R (8.31 J/molK) and the temperature (75°C = 75 + 273 = 348 K).
Let's consider the reaction:
Ni(g) + 4CO(g) → Ni(CO)4(g)
From the reaction, we can see that 4 moles of CO react with 1 mole of Ni to form 1 mole of Ni(CO)4. Therefore, the change in moles (Δn) is equal to (1 - 0) = 1.
Now, let's calculate the volume change during the reaction. Since all gases are ideal, we can use the ideal gas equation.
Rearranging the equation to solve for V:
V = (nRT) / P
We know the initial conditions, where n = 0 (before any reaction) and P = initial pressure. However, since we are only interested in the volume change, we can disregard the initial values and consider the final values, where n = 1 mole and P = final pressure.
Therefore, we can calculate ΔV as:
ΔV = (nRT / P) - (0RT / P)
= (1 mol * 8.31 J/molK * 348 K) / P
Now, we need to find the final pressure (P). To do this, we can use the ideal gas law equation again, but this time considering the final moles.
According to the reaction, 1 mole of Ni and 4 moles of CO react to form 1 mole of Ni(CO)4. So, the total number of moles after the reaction is (0 + 4) = 4 mole.
Rearranging the ideal gas equation for P:
P = (nRT) / V
Substituting the values, we get:
P = (4 mol * 8.31 J/molK * 348 K) / ΔV
Now, we have the final pressure. To calculate the work done (W), we use the formula:
W = -PΔV
Substituting the values, we get:
W = -[(4 mol * 8.31 J/molK * 348 K) / ΔV] * ΔV
= -4 * 8.31 J/molK * 348 K
Calculating this expression will give you the answer.
sorry, I pasted the wrong reaction. thanks for the help
Did you give the reaction below?
yeah its here
HNO3(aq) + NaOH(s) -----> NaNO3(aq) + H2O(l)