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1 month ago

The equilibrium constant k for the synthesis of ammonia is 6.8x105 at 298 k. what will k be for the reaction at 375 k?

Chemistry

2 answers:

eduard [2.7K]1 month ago

8 0

Missing in your question:

ΔH= -92.22 KJ.mol^-1 and the reaction equation is:
N2(g) +3H2(g)⇄2NH3(g)
Using this formula:
㏑(K2/K1) = -ΔH/R (1/T2 - 1/T1)
with ΔH= 92.22 KJ.mol^-1 & K2= X & K1= 6.8X10^5 & T1= 298 K &
T2=375 K & R constant= 8.314
Plugging in these values allows us to find K2.
㏑(X/6.8x10^5) = 92.22/8.314 (1/375 - 1/298)
Thus, X= 6.75X10^5
Hence, K2 = 6.75X10^5

Alekssandra [3K]1 month ago

6 0

The equilibrium constant at 375 K is K = 326. The chemical reaction involved is: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) with a ΔH of -92.22 kJ/mol. For the temperature values, T₁=298 K and T₂=375 K. Converting ΔH, we get -92.22 kJ/mol which equals -92220 J/mol. The gas constant R is 8.314 J/K·mol. K₁ is 6.8·10⁵ while K₂ remains to be determined. Using the van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁), we have ln(K₂/6.8·10⁵) = 92220 J/mol / 8.314 J/K·mol · (1/375K - 1/298K). This results in ln(K₂/6.8·10⁵) = 11092.13 · (0.00266 - 0.00335). Thus, ln(K₂/6.8·10⁵) results in -7.64. Solving for K₂ gives K₂/680000 = 0.00048, leading to K₂ being approximately 326.4.

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1 month ago

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18 days ago

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Answer:

Quantity of $H_{2}O$ generated will be reduced to fifty percent of its initial amount.

Explanation:

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2 months ago

A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane

Tems11 [2777]

Explanation:

The rate at which gases effuse is inversely related to the square root of their molar masses.

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3 0

1 month ago

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