On a certain hot day the atmospheric pressure was 1 atm and the temperature was 37°C. On this day 1 L sample of atmospheric gases was collected and analyzed. It was found to contain among other gases 9.38 x 10-3 mol of O2 gas, 2.46 x 10-8 mol of H2 gas and 1.79 x 10-4 mol of water vapor. If you put these amounts of the three gases in a sealed 1L tank at 1 atm and 37°C, will the combustion of hydrogen gas be spontaneous at these conditions?

Answer:

Yes, the combustion of hydrogen is spontaneous from the given conditions.

Explanation:

Let look at the reaction:

[tex]2H_2_{(g)} + O_2_{(g)} \to 2H_2O_{(g)}[/tex]

[tex]K_p = \dfrac{[P(H_2O)]^2}{[P(H_2)^2\times P(O_2)]} ----- (1)[/tex]

The total pressure of the tank = 1 atm

[tex]\mathtt{Pressure \ of \ the \ gas = moles \ fraction \of \ gas \times total \ pressure}[/tex]

Thus, pressure of [tex]O_2[/tex] gas = [tex]\dfrac{9.38 \times 10^{-3} }{(9.38 \times 10^{-3} + 2.46 \times 10^{-8} + 1.79 \times 10^{-4}} \times 1 atm[/tex]

[tex]P(O_2) = 0.9813 atm[/tex]

Pressure of [tex]H_2O[/tex] gas =[tex]\dfrac{1.78 \times 10^{-4}}{9.38 \times 10^{-3} + 2.46 \times 1)^{-8} + 1.78 \times 10^{-4}} \times 1 \ atm[/tex]

Pressure of H_2 gas = [tex]2.573 \times 10^{-6} \ atm[/tex]

From equation (1);

[tex]K_p = \dfrac{(0.0187)^2}{(2.573 \times 10^{-6})^2 \times (0.913) } \\ \\ K_p = 5.383 \times 10^7[/tex]

So, applying the equation:

[tex]\Delta G^0 = -RT In K_p \\ \\ = -8.314 \times (300)K \times IN(5.383 \times 10^7) \\ \\ = -45880.109 J \\ \\ = -45.800 \ kJ\\ \\[/tex]

Thus;

[tex]\Delta G^0 < 0[/tex]

Thus, Yes the combustion of hydrogen is spontaneous from the above conditions. From thermodynamics, we realize that assuming ∆G is negative, the reaction is spontaneous. Again we understand that ∆G=∆H-T∆S. However, for combustion, water will definitely be formed.

Finally, we can conclude that;

∆H is negative;

∆S is positive; &

∆G is also negative.

Thus, from above, we conclude combustion of hydrogen is spontaneous.

The combustion of hydrogen gas with the oxygen gas for the formation of the water vapor is a spontaneous reaction.

To know about the condition that the given reaction is spontaneous or not we will use the below equation and must get the negative value:

ΔG° = -RTlnKp, where

R = universal gas constant

T = temperature

Kp = partial pressure constant

Firstly we have to calculate the value of Kp.

Given chemical reaction is:

2H₂(g) + O₂(g) → 2H₂O(g)

Value of Kp for the given reaction is:

Kp = [p(H₂O)]² / [p(H₂)]².[p(O₂)]

Partial pressure will be calculated as:

p = mole fraction × total pressure

Given that total pressure of gas = 1atm

p(H₂O) = (1.79×10⁻⁴ / 1.79×10⁻⁴+2.46×10⁻⁸+9.38×10⁻³) × 1 = 0.0187

p(H₂) = (2.46×10⁻⁸ / 1.79×10⁻⁴+2.46×10⁻⁸+9.38×10⁻³) × 1 = 2.573 × 10⁻⁶

p(O₂) = (9.38×10⁻³ / 1.79×10⁻⁴+2.46×10⁻⁸+9.38×10⁻³) × 1 = 0.913

Putting all these values on the equation of Kp, we get

Kp = (0.0187)² / (2.573 × 10⁻⁶)²(0.913) = 5.383 × 10⁷

Now putting values in the first equation we get,

ΔG° = -(8.314)(300)ln(5.383 × 10⁷)

ΔG° = -45,800 kJ

Hence, given reaction is spontaneous.

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