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3 years ago

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The reaction of methane with water to form carbon dioxide and hydrogen is nonspontaneous at 298 K. At what temperature will this

system make the transition from nonspontaneous to spontaneous? The data refer to 298 K. CH4(g) + 2H2O(g) CO2(g) + 4H2(g) Substance: CH4(g) H2O(g) CO2(g) H2(g) ΔH°f (kJ/mol): –74.87 –241.8 –393.5 0 ΔG°f (kJ/mol): –50.81 –228.6 –394.4 0 S°(J/K·mol): 186.1 188.8 213.7 130.7

1 answer:

KIM [24]3 years ago

8 0

Answer:

The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.

Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.

Explanation:

CH4(g) + 2H2O(g) ⇆ CO2(g) + 4H2(g)

CH4(g) H2O(g) CO2(g) H2(g) ΔH°f (kJ/mol): –74.87 –241.8 –393.5 0

ΔG°f (kJ/mol): –50.81 –228.6 –394.4 0

S°(J/K·mol): 186.1 188.8 213.7 130.7

ΔG<0 to be spontaneous

ΔG = ΔH- TΔS <0

ΔH = ∑nΔH(products) - ∑nΔH(reactant)

ΔH = (-393.5) - (–74.87 + 2*–241.8)

ΔH = 164.97 kJ = 164970 J

ΔS = ∑nΔS(products) - ∑nΔS(reactant)

ΔS = (213.7 + 4*130.7) - (186.1 + 2*188.8)

ΔS = 172.8 J

0 > 164970 J - T* 172.8 J

-164970 J > - T* 172.8 J

954.7< T

The system makes the transition from nonspontaneous to spontaneous at a temperature of 954.7 K.

Under 954.7 K the reaction is nonspontaneous; more than 954.7 K is the reaction spontaneous.

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Explanation:

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Solution:

As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....

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<h3>2nd Method:</h3>

Assuming that the gas is acting ideally, hence, applying ideal gas equation.

P V = n R T ∴ R = 0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹

Solving for n,

n = P V / R T

Putting values,

n = (1 atm × 87.6 L)/(0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15K)

n = 3.91 moles

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87.6 L of Neon gas will contain 3.91 moles at standard temperature and pressure.

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