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Is the following reaction a combustion reaction or just a single displacement? Could be both?

$$\ce{H2 + CuO -> Cu + H2O}$$

A friend of mine claims it is a combustion reaction because it is exothermic and a redox reaction.

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    $\begingroup$ No, combustion is a loose term which implies a visible flame. $\endgroup$
    – AChem
    Commented May 27, 2020 at 14:46
  • $\begingroup$ Think about what all is required for a combustion reaction (think formulas) $\endgroup$ Commented May 27, 2020 at 14:55
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    $\begingroup$ @kskinnerx16, There is no formula for combustion if you are thinking of carbon dioxide and water as products. All you need is a (usually) visible flame to call it a combustion process. Sodium can "burn" in chlorine, right? $\endgroup$
    – AChem
    Commented May 27, 2020 at 14:59
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    $\begingroup$ @M.Farooq Certainly. Perhaps we need more context from OP, as I took this as a fairly basic level question (identifying reaction types). $\endgroup$ Commented May 27, 2020 at 15:01
  • $\begingroup$ For the record, so nano-related thermites are not associated with 'combustion reaction happens quickly, producing heat, and usually light and fire. " My advice, do not even attempt one for safety reasons. $\endgroup$
    – AJKOER
    Commented May 27, 2020 at 17:41

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By the question asked, I assume OP is in high school of freshman in college. Thus, I'd use this general description of combustion as a starting point:

Combustion reactions are common and very important. Combustion means burning, usually in oxygen but sometimes with other oxidants such as fluorine. A combustion reaction happens quickly, producing heat, and usually light and fire. Combustion describes how the reaction happens, not the reactants and products. Chemists as early as Lavoisier suggested that people get their energy from combustion-like reactions, but even though the products and reactants are the same when you burn food in a fire and in your body, the way it happens is different. In a combustion reaction, the thing that burns (the reactant that isn't $\ce{O2}$ or $\ce{F2}$) is called the fuel. Combustion reactions are a type of redox reaction.

As M. Farooq pointed out a combustion reaction happens quickly, producing heat, and usually light and fire. For example, lets look at combustion reaction of an alkene (a hydrocarbon). If it is a complete combustion, the fire have a blue flame:

$$\ce{C_nH_{2n} + $\frac{3n}{2}$ O2 -> nCO2 + n H2O}$$

If it is a partial combustion, it can have a multiple $\ce{C}$ compounds as products, and have a yellow flame due to presence of elemental $\ce{C}$:

$$\ce{C_nH_{2n} + x O2 -> m C + p CO + $(n-m-p)$CO2 + n H2O}$$

where $x = \frac{2(n-p-m) +p}{2} = \frac{2n-2p-2m +p}{2} = \frac{2n-p-2m)}{2}$. In your reaction would not produce fire and it didn't use either oxygen or other oxidants ($\ce{CuO}$ is not that type of oxidant). It is true that the reaction is a redox reaction.

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    $\begingroup$ @KareemAli, he means CuO is not a gaseous oxidizer, which is typically needed in a combustion reaction. $\endgroup$
    – AChem
    Commented May 27, 2020 at 16:24
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    $\begingroup$ LOL - The reaction is also a hard way to get hydrogen to oxidize. $\endgroup$
    – MaxW
    Commented May 27, 2020 at 16:30
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    $\begingroup$ @ M. Farooq: Man, you read my mind! :-) $\endgroup$ Commented May 27, 2020 at 16:37
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    $\begingroup$ @MaxW, this reaction was historically very important in analytical chemistry. "Copper oxide (CuO) is used extensively in elemental analysis for the oxidation of light hydrocarbons and carbon monoxide into carbon dioxide and of hydrogen into water" $\endgroup$
    – AChem
    Commented May 27, 2020 at 17:27
  • $\begingroup$ Actually, with nano CuO, a remarkably exothermic reaction with hydrogen is expected per my cited Science Direct source. $\endgroup$
    – AJKOER
    Commented May 27, 2020 at 17:30
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Your friend is right, it 'can' produce actually an extraordinary combustion/thermite reaction.

In fact, for example, one can expect extreme temperatures when mixing fine metal powder (for example Aluminum dust) with nano-CuO. Per this June 2019 source from Science Direct, to quote from the abstract:

Nanothermites (metal oxide/metal) can offer tremendously exothermic self sustained reactions. CuO is one of the most effective oxidizers for naonothermite applications. This study reports on two prospectives for the manufacture of CuO nanoparticles....The superior performance of colloidal CuO particles was correlated to their steric stabilization with employed organic solvent. This is the first time ever to report on fabrication, isolation, and integration of stablilized colloidal nanothermite particles into energetic matrix where intimate mixing between oxidizer and metal fuel could be achieved

And from the body of the work:

CuO/Al is one of the most effective thermite mixtures in terms of combustion temperature, heat output, and gasous products [9,10]. CuO/Al can offer an adiabatic flame temperature of 5718 K with heat of reaction of about 4100 J/g, and a reasonable amount of gaseous products (0.345 g gas/g thermite). Furthermore the formed Cu metal (resulted from thermite reaction) can be transformed into gas due to its low boiling point (1000 °C) [11].

Working with gaseous hydrogen in place of Aluminum is expected to be less exothermic, but more likely, I suspect, an explosion hazard. For support, based on this 1940 article, for my supposition based on the known combination reaction of H2 with O2 in the catalytic presence of Cu metal (likely with a Cu2O layer). This likely implies a possible radical-induced hydrogen atom acting on oxygen (nano-CuO) pathway.

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  • $\begingroup$ I do not know who has a problem with my cited source (Science Direct) and is further not acquainted with the activity of nano-oxides, but I will kindly provide educational references. $\endgroup$
    – AJKOER
    Commented May 27, 2020 at 17:35
  • $\begingroup$ So, nano-related thermites are not associated with 'combustion reaction happens quickly, producing heat, and usually light and fire. " ? My advice, do not even attempt one for safety reasons. $\endgroup$
    – AJKOER
    Commented May 27, 2020 at 17:40

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