Answer:
Explanation:
For the reaction ,
2CH₃OH + 3O₂ → 2CO₂ + 4 H₂O
For the above reaction ,
the change in enthalphy is calculated as
Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)
In case the compound is in its standard state , enthalphy of formation is zero
Hence ,
for the above reaction ,
ΔHrxn =( 2 * Δ H° (CO₂ ) + 4 * Δ H° (H₂O )) - [ ( 2 *Δ H°CH₃OH ) + (3 * Δ H° O₂ )]
Δ H° (CO₂ ) = -393.5kJ /mol
Δ H° (H₂O ) = - 241.8 kJ /mol
Δ H°CH₃OH = -239.2kJ /mol
Δ H° O₂ = 0
putting the corresponding values ,
ΔHrxn =( 2 * -393.5kJ /mol + 4 *- 241.8kJ /mol) - [ ( 2 *-239.2kJ /mol ) + (3 *0 )
ΔHrxn = -1275.8 kJ /mol
Moles of methanol,
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
w = 125 g
as we know ,
m = 32 g /mol
n = 125 g / 32 g /mol = 3.906 mol
From , the reaction , 2 mol produces -1275.8 kJ /mol heat ,
Now using unitary method ,
1 mol produces = -1275.8 kJ /mol / 2 heat ,
3.906 mol produces = -1275.8 kJ /mol / 2 * 3.906 heat
3.906 mol produces = 249.7 kJ
