Answer:
3.94 L
Explanation:
From the question given above, the following data were obtained:
Mass of O₂ = 5.62 g
Volume of O₂ =?
Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:
Mass of O₂ = 5.62 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mole of O₂ =?
Mole = mass / molar mass
Mole of O₂ = 5.62 / 32
Mole of O₂ = 0.176 mole
Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.
Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP
Answer:
-3135.47 kJ/mol
Explanation:
Step 1: Write the balanced equation
C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)
Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)
We will use the following expression.
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpies of formation
p: products
r: reactants
ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))
ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol
ΔH°r = -3135.47 kJ
Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.
Answer and explanation:
The relative rates of free radical halogenation is in the order of,
F₂ (10⁸) > Cl₂ (1) > Br₂ (10⁻¹¹) > I₂ (10⁻²²)
The above order also show a decreasing reactivity from left to right
Hence, reaction of fluorine with alkanes is highly reactive and it is too difficult to control.
The reaction with Cl₂ is moderately fast, while with Br₂ is slow and with I₂ is too slow reaches to equilibrium.
Hence in general we don't prefer radical fluorination of alkanes.
