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Andrews [41]
3 years ago
12

What is the volume of 1.0 mole of hydrogen gas at standard temperature and pressure ?

Chemistry
2 answers:
Reil [10]3 years ago
8 0

22.4 liters

Molar volume, or volume of one mole of gas , depends on pressure and temperature, and is 22.4 liters - at 0 °C (273.15 K) and 1 atm (101325 Pa), or STP (Standard Temperature and Pressure), for every gas which behaves similarly to an ideal gas.

mark brainliest please

finlep [7]3 years ago
6 0

at standard it is 22.4 liters

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45 m/s north

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6. Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions Ce4+, Ce3+, Fe3+, an
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ΔG° = -80.9 KJ

Assuming this reaction takes place at room temperature (25 °C):

K=1.53x10^{14}

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First of all one should look up the reduction potentials for the species envolved:

Ce^{4+} + e→Ce^{3+}         E°red=1.61V

Fe^{3+} + e→Fe^{2+}         E°red=0.771V

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Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. <u>Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).</u>

1)  Ce^{4+} reduces while  Fe^{2+} oxidates

  (oxidation)               Fe^{2+}→Fe^{3+} + e          E°oxi=-0.771V

  (reduction)               Ce^{4+} + e→Ce^{3+}         E°red=1.61V

  (overall equation)    Fe^{2+}+Ce^{4+}→Ce^{3+}+Fe^{3+} E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v

The cell potential can also be calculated as the cathode potencial minus the anode potential:

E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v

3) Gibbs free energy and Equilibrium constant

ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.

ΔG°=-nFE°=-1*96500*0.839

ΔG° = - 80963 J = -80.9 KJ

The Nerst equation gives us the relation of chemical equilibrium and Electric potential.

E=E°-\frac{RT}{nF} Ln Q

Where 'R' is the molar gas constant (8.314 J/mol)

It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:

E°=\frac{RT}{nF} Ln K

Isolating for 'K' gives:

K=e^{\frac{nFE^{o} }{RT} }

This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):

K=1.53x10^{14}

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