Calculate the change in enthalpy when 52.0 g of solid chromium at 25°C and 1 atm pressure is oxidized. (H°f for Cr2O3(s) is –1135 kJ/mol.)

4Cr(s) + 3O2(g)  2Cr2O3(s)

a. –1135 kJ
b. –284 kJ
c. –568 kJ
d. +1135 kJ
e. +568 kJ

To calculate the change in enthalpy when 52.0 g of solid chromium is oxidized, we need to use the given equation and the molar mass of chromium.

1. Calculate the number of moles of chromium:
Molar mass of chromium (Cr) = 52.0 g/mol

2. Use the molar mass of chromium to calculate the number of moles:
Number of moles of chromium = (52.0 g) / (molar mass of chromium)

3. Determine the change in enthalpy using the stoichiometry of the balanced equation:
Change in enthalpy = (Number of moles of chromium) * (ΔHf for Cr2O3)

4. Substitute the values into the equation:
Change in enthalpy = (Number of moles of chromium) * (-1135 kJ/mol)

5. Calculate the change in enthalpy:
Change in enthalpy = (Number of moles of chromium) * (-1135 kJ/mol)

Now you can perform the calculations to find the change in enthalpy.

To calculate the change in enthalpy (ΔH) when solid chromium is oxidized, we need to use the balanced chemical equation and the given enthalpy of formation (ΔH°f) for Cr2O3(s).

From the balanced chemical equation: 4Cr(s) + 3O2(g) → 2Cr2O3(s)

The stoichiometric coefficient in front of Cr2O3(s) is 2, meaning that 2 mol of Cr2O3(s) are formed for every 1 mol of the reaction.

Since the given enthalpy of formation (ΔH°f) for Cr2O3(s) is -1135 kJ/mol, this means that the enthalpy change for the formation of 2 mol of Cr2O3(s) is -1135 kJ/mol x 2 = -2270 kJ.

However, we need to find the enthalpy change for the formation of 52.0 g of Cr2O3(s).

First, we need to calculate the molar mass of Cr2O3:
2(52.0 g of Cr) + 3(16.0 g of O) = 104.0 g of Cr2O3

Next, we can find the number of moles of Cr2O3:
moles = mass / molar mass
moles = 52.0 g / 104.0 g/mol
moles = 0.5 mol

Finally, we can calculate the change in enthalpy (ΔH):
ΔH = ΔH°f x moles
ΔH = -2270 kJ/mol x 0.5 mol
ΔH = -1135 kJ

Therefore, the change in enthalpy when 52.0 g of solid chromium is oxidized is -1135 kJ.

So, the correct answer is option a. -1135 kJ.

The dH rxn = -1135 kJ/mol x 2 mol = ?

That is for 4*52 g Cr.
-1135*2 x (52/4*52) = ?