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

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Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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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...