Given the thermochemical equations below, What is the standard heat of formation of CuO(s)? 2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ

-130KJ is the standard heat of formation of CuO.

Explanation:

The standard heat of formation or enthalpy change can be calculated by using the formula:

standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation

Data given:

Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ

2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ

CuO + Cu ⇒ Cu2O (-11.3 KJ)      ( Formation of Cu2O)

When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.

Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ

So standard heat of formation of  formation of Cu0 as:

Cu + 1/2 02 ⇒ CuO

putting the values in the equation

ΔHf = ΔH1 + ΔH2     (ΔH1 + ΔH2  enthalapy of reactants)

heat of formation = -11.3 + (-119.35)

                            = - 130.65kJ

-130.65 KJ is the heat of formation of CuO in the given reaction.

The standard heat of formation of CuO is -310.5 kJ.

Explanation:

To get the standard heat of formation, we have to multiply the second equation by 2 and reverse it and add it to the first equation, we will get the required compound.

2 Cu₂O(s) + O₂(g) → 4 CuO(s) ∆H° = -287.9 kJ 1

Cu₂O(s) → CuO(s) + Cu(s)        ∆H° = 11.3 kJ 2

Now multiplying the equation 2 by 2 and reversing it, we will get,

2 CuO(s) + 2 Cu(s) → 2 Cu₂O(s)     ∆H° = -22.6 kJ 3

Now adding eq. 3 with eq. 1 we will get the formation of CuO as,

2 Cu(s) + O₂(g) → 2 CuO(s) ∆H° = -22.6 kJ - 287.9 kJ = -310.5 kJ

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b