If I heat ammonia gas to 250 degrees celsius, will it decompose into its basic components - N2 and H2? What temperature is required for the reaction to occur?
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$\begingroup$ Only on far higher temperature. Afaik 250 is far not enough for that, maybe if you have some catalyzers, but even so is it not sure. As far I know, the composition of NH3 is exotherm, at least on this temperature. The problem of the creation of ammonia is that the triple bonds of the N2 are very hard to dismantle. This is why ammonia creation requires well-controlled temperature (more than 250), pressure and catalyzers, in a many-step process. But I am not very sure in that. $\endgroup$– peterhCommented Mar 12, 2020 at 12:28
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The equation of the reaction is : $$\ce{N_2(g) + 3 H_2(g) <=> 2 NH_3(g)} $$
The equilibrium constant for this reaction have been summarized in the Chemistry Data Book, by J. G. Stark and H.G. Wallace, Jon Murray, London, 2006. At $\pu{400 °C},$ $K_p$ is $\pu{40.7 atm^{-2}}.$ At $\pu{500 °C},$ $K_p$ is equal to $0.035$. So, at $\pu{250 °C},$ $\ce{NH3}$ is not significantly decomposed.
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See Highly efficient decomposition of ammonia using high-entropy alloy catalysts which proposes highly efficient ammonia decomposition using a novel high-entropy alloy (HEA) catalysts made of earth abundant elements, namely, quinary CoMoFeNiCu nanoparticles.
However, the reported temperature is high (going up to 2000–2300 K).
Another source notes:
To date, the most effective catalyst for ammonia decomposition consists of ruthenium particles supported on carbon nanotubes (CNT) due to their high conductivity (6353 molH2 mol−1Ruh−1 at 430 °C) [2, 17]. The low temperature activity can be further improved by the addition of an electron donating promoter such as cesium (7870 molH2 mol−1Ruh−1 at 370 °C) [3, 17, 19, 20].
So, to answer your question, not at the temperature you indicated (250 degrees Celsius).
