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Sagot :
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.
How do we know reaction is spontaneous?
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.
To know more about spontaneous reaction, visit the below link:
https://brainly.com/question/24376583
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