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Drupady [299]
3 years ago
6

The normal boiling point of methanol is 64.7∘c and the molar enthalpy of vaporization of 71.8kj/mol. the value of δs when 1.11 m

ol of ch3oh(l) vaporizes at 64.7∘c is ________ j/k.
Chemistry
1 answer:
maxonik [38]3 years ago
6 0
The answer is 236.5 J/K

According to Δ G formula:

ΔG = ΔH - TΔS

when ΔG is the change in free energy (KJ)

and ΔH is the change in enthalpy (KJ)= ΔHvap * moles

                                                              = 71.8 KJ/mol * 1.11 mol
                                                             
                                                              =   79.7 KJ

and T is the absolute temperature (K)= 64 °C + 273°C = 337 K

Δ S is the change in entropy  KJ/K

by substitution:

when at equilibrium ΔG = 0 

∴ΔS = ΔH / T

       =79.7 KJ/ 337 K

     = 0.2365 KJ/K

     = 236.5 J/K


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Marina CMI [18]

Answer:

9.1 mol

Explanation:

The balanced chemical equation of the reaction is:

CO (g) + 2H2 (g) → CH3OH (l)

According to the above balanced equation, 2 moles of hydrogen gas (H2) are needed to produce 1 mole of methanol (CH3OH).

To convert 36.7 g of hydrogen gas to moles, we use the formula;

mole = mass/molar mass

Molar mass of H2 = 2.02g/mol

mole = 36.7/2.02

mole = 18.17mol

This means that if;

2 moles of H2 reacts to produce 1 mole of CH3OH

18.17mol of H2 will react to produce;

18.17 × 1 / 2

= 18.17/2

= 9.085

Approximately to 1 d.p = 9.1 mol of methanol (CH3OH).

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<u><em> calculation</em></u>

step 1: write the equation  for reaction

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