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Enter a redox reaction equation to balance it and calculate the reducing and oxidizing agents.
14 sie 2020 · Because H 2 is a good reductant and O 2 is a good oxidant, this reaction has a very large equilibrium constant (K = 2.4 × 1047 at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form O 2 and H 2, is very small: K′ = 1 / K = 1 / (2.4 × 1047) = 4.2 × 10 − 48.
To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. The balanced equation will appear above. Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored.
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16 kwi 2015 · The hydroxyl group of alcohols is normally a poor leaving group. However, when treated with strong acid, R-OH is converted into R-OH 2 (+) and H 2 O is a much better leaving group. With tertiary alcohols, H 2 O can then leave, resulting in a carbocation.
5 lut 2015 · 7 Answers. Sorted by: 11. +50. The question: Will NaOH + K form NaOK or KOH? This is a redox reaction regardless of what species K would react with. The answer is two-fold: (a) Having potassium added to an aqueous solution of sodium hydroxide; and (b) Having potassium added to molten sodium hydroxide, therefore no water present (or trace amount).
23 wrz 2022 · \[\Delta H_{rxn}=\left [ \left ( 2\cancel{mol} \right )\left ( -92.3kJ/\cancel{mol} \right )+\left ( 1\cancel{mol} \right ) \left ( 0kJ/\cancel{mol} \right )\right ]-\left [ \left ( 2\cancel{mol} \right )\left ( -36.3kJ/\cancel{mol} \right )+\left ( 1\cancel{mol} \right ) \left ( 0kJ/\cancel{mol} \right )\right ]\nonumber \]
