Chemistry, 23.07.2019 21:00 AliceYT

The normal boiling point of methanol is 64.7 °c and the molar enthalpy of vaporization if 71.8 kj/mol. the value of ̇s when 1.11 mol of ch3oh (l) vaporizes at 64.7 °c is j/k.

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Answer from: Bloom247

The change in entropy of the system with the vaporization of methanol has been 236 J/K.

The boiling point has been defined as the amount of heat required to turn the liquid into steam. The molar enthalpy of vaporization has been the amount of heat required to convert 1 mole of water to steam.

Computation for entropy

The value of entropy ( $\Delta S^\circ$ ) has been given as:

$\Delta S^\circ=\dfrac{n \Delta H^\circ}{T}$

Where, the moles of methanol has been, $n=1.11\;\rm mol$

The enthalpy of vaporization of methanol ( $\Delta H^\circ$ ),

$\Delta H^\circ =71.8\;\text{kJ/mol}\\
\Delta H^\circ =71800\;\text{J/mol}$

The temperature of the system (T) has been,

$T=64.7^\circ C\\
T=64.7\;+\;27.15\;\text K\\
T=337.7 \;\text K$

Substituting the values for the entropy ( $\Delta S^\circ$ ) of the reaction:

$\Delta S^\circ =\dfrac{1.11\;\text{mol}\;\times\;71800\;\text{J/mol}}{337.7\;\text K} \\
\Delta S^\circ =236\;\rm J/K$

The change in entropy of the system with the vaporization of methanol has been 236 J/K.

Learn more about change in entropy, here:

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Answer from: hudsonlilliana

When ΔG° = ΔH° - TΔS° = 0 since it is at the normal boiling point, so:

we will use this formula:

ΔS° = ΔH°/T and this formula for only 1 mol CH3OH but when we have n (number of moles) the formula will be:

ΔS° = n* ΔH° / T

when n is the number of moles of CH3OH = 1.11 moles

and Δ H° = the molar enthalpy of vaporization = 71.8 Kj/mol

T is the temperature in Kelvin = 64.7°C +273 = 337.7 K

so by substitution:

∴ ΔS° = 1.11 moles * 71800 J/mol / 337.7

= 236 J/K

∴the answer is 236 J/K

Answer from: Quest

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

Answer from: Quest

Spontaneity tell us the direction of the reaction, but not how fast it goes. gibbs free energy brings together enthalpy, entropy and temperature in one easy formula. g = h - ts. and, finally the sign of delta g is always negative for a spontaneous reaction.

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The normal boiling point of methanol is 64.7 °c and the molar enthalpy of vaporization if 71.8 kj/mo...