If 15.0 kJ are released when 1.40 g of F2 reacts with an excess of SiH4, complete the thermochemical equation below.
SiH4(g) + 2 F2(g) SiF4(g) + 4 HF(g)
ÄHrxn=?kJ
Wouldn't it be
15.0 kJ x (4 x atomic mass F/1.50) = ??
Check my thinking.
814
To find the enthalpy change (ΔHrxn) for the reaction, we need to calculate the moles of F2 used in the reaction.
First, we convert the mass of F2 to moles:
1.40 g F2 * (1 mol F2/38.0 g F2) = 0.037 mol F2
From the balanced equation, we see that the stoichiometric coefficient of F2 is 2, so to find the moles of SiH4 that reacted, we can use the mole ratio between SiH4 and F2:
0.037 mol F2 * (1 mol SiH4/2 mol F2) = 0.0185 mol SiH4
Now, we can use the given energy released (15.0 kJ) and the moles of SiH4 to calculate the enthalpy change:
ΔHrxn = Energy released / moles of SiH4
ΔHrxn = 15.0 kJ / 0.0185 mol
ΔHrxn ≈ -810.8 kJ/mol
Since the given energy is negative (released), the enthalpy change for the reaction is approximately -810.8 kJ/mol.
To find the value of ΔHrxn, the change in enthalpy for the reaction, we can use the concept of stoichiometry and the given information about the energy released.
First, let's determine the number of moles of F2 involved in the reaction. We know that the mass of F2 is 1.40 g, so we need to convert this mass to moles.
To do this, we use the molar mass of F2, which is 38.0 g/mol.
moles of F2 = mass of F2 / molar mass of F2
moles of F2 = 1.40 g / 38.0 g/mol
Now, with the number of moles of F2, we can use the stoichiometric coefficients of the balanced equation to find the value of ΔHrxn.
From the balanced equation:
2 moles of F2 produce 15.0 kJ
So, we can set up a ratio to find the value of ΔHrxn:
15.0 kJ / 2 moles of F2 = ΔHrxn / moles of F2
Now substitute the values we have:
15.0 kJ / 2 moles = ΔHrxn / (1.40 g / 38.0 g/mol)
Simplifying the equation:
ΔHrxn = (15.0 kJ / 2) * (1.40 g / 38.0 g/mol)
Calculating this expression will give us the value of ΔHrxn.