ch3oh + co


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Word Equation

Methanol + Carbon Monoxide = Acetic Acid

One mole of

Methanol [CH3OH]

and one mole of

Carbon Monoxide [CO]

combine lớn sườn one mole of

Acetic Acid [CH3COOH]


  • Methanol - CH3OH

    Methyl Hydroxide Hydroxymethane Monohydroxymethane Colonial Spirit Columbian Spirit Spirit Of Wood Pyroxylic Spirit Methylol CH3OH Molar Mass CH3OH Oxidation Number

  • Carbon Monoxide - CO


Thermodynamics of the reaction can be calculated using a lookup table.

Is the Reaction Exothermic or Endothermic?

CH3OH (g methanol)1 mol-201.08304 kJ/mol201.08304 kJ
CO (g)1 mol-110.54128 kJ/mol110.54128 kJ
CH3COOH (l acetic acid)1 mol-484.13064 kJ/mol-484.13064 kJ
ΣΔH°f(reactants)-311.62432 kJ
ΣΔH°f(products)-484.13064 kJ
ΔH°rxn-172.50632 kJ

ΣΔH°f(reactants) > ΣΔH°f(products), sánh CH3OH + CO = CH3COOH is exothermic (releases heat).

Is the Reaction Exoentropic or Endoentropic?

ΔS = Sproducts - Sreactants. If ΔS < 0, it is exoentropic. If ΔS > 0, it is endoentropic.

CH3OH (g methanol)1 mol239.70136 J/(mol K)-239.70136 J/K
CO (g)1 mol197.9032 J/(mol K)-197.9032 J/K
CH3COOH (l acetic acid)1 mol159.8288 J/(mol K)159.8288 J/K
ΣΔS°(reactants)437.60456 J/K
ΣΔS°(products)159.8288 J/K
ΔS°rxn-277.77576 J/K

ΣΔS°(reactants) > ΣΔS°(products), sánh CH3OH + CO = CH3COOH is exoentropic (decrease in entropy).

Is the Reaction Exergonic or Endergonic?

ΔG = Gproducts - Greactants. If ΔG < 0, it is exergonic. If ΔG > 0, it is endergonic.

CH3OH (g methanol)1 mol-162.42288 kJ/mol162.42288 kJ
CO (g)1 mol-137.27704 kJ/mol137.27704 kJ
CH3COOH (l acetic acid)1 mol-389.9488 kJ/mol-389.9488 kJ
ΣΔG°(reactants)-299.69992 kJ
ΣΔG°(products)-389.9488 kJ
ΔG°rxn-90.24888 kJ

ΣΔG°(reactants) > ΣΔG°(products), sánh CH3OH + CO = CH3COOH is exergonic (releases energy).

Reaction Expressions

Kc or Q = ( [CH3COOH] ) / ( [CH3OH] [CO] )

(assuming all reactants and products are aqueous. substitutue 1 for any solids/liquids, and Psubstance for gases.)

rate = -(Δ[CH3OH] / Δt) = -(Δ[CO] / Δt) = (Δ[CH3COOH] / Δt)

(assuming constant volume in a closed system and no accumulation of intermediates or side products)


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  • Use uppercase for the first character in the element and lowercase for the second character. Examples: Fe, Au, Co, Br, C, O, N, F.
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Balance CH3OH + CO = CH3COOH Using the Algebraic Method

To balance the equation CH3OH + CO = CH3COOH using the algebraic method step-by-step, you must have experience solving systems of linear equations. The most common methods are substitution/elimination and linear algebra, but any similar method will work.

  1. Label Each Compound With a Variable

    Label each compound (reactant or product) in the equation with a variable lớn represent the unknown coefficients.

    a CH3OH + b CO = c CH3COOH

  2. Create a System of Equations

    Create an equation for each element (C, H, O) where each term represents the number of atoms of the element in each reactant or product.

    C: 1a + 1b = 2c H: 4a + 0b = 4c O: 1a + 1b = 2c

  3. Solve For All Variables

    Use substitution, Gaussian elimination, or a calculator lớn solve for each variable.

    • 1a + 1b - 2c = 0
    • 4a - 4c = 0
    • 1a + 1b - 2c = 0

    Use your graphing calculator's rref() function (or an online rref calculator) lớn convert the following matrix into reduced row-echelon-form:

    [ 1 1 -2 0] [ 4 0 -4 0] [ 1 1 -2 0]

    The resulting matrix can be used lớn determine the coefficients. In the case of a single solution, the last column of the matrix will contain the coefficients.

    Simplify the result lớn get the lowest, whole integer values.

    • a = 1 (CH3OH)
    • b = 1 (CO)
    • c = 1 (CH3COOH)
  4. Substitute Coefficients and Verify Result

    Count the number of atoms of each element on each side of the equation and verify that all elements and electrons (if there are charges/ions) are balanced.

    CH3OH + CO = CH3COOH

    Reactants Products

    Since there is an equal number of each element in the reactants and products of CH3OH + CO = CH3COOH, the equation is balanced.

Balance CH3OH + CO = CH3COOH Using Inspection

The law of conservation of mass states that matter cannot be created or destroyed, which means there must be the same number atoms at the over of a chemical reaction as at the beginning. To be balanced, every element in CH3OH + CO = CH3COOH must have the same number of atoms on each side of the equation. When using the inspection method (also known as the trial-and-error method), this principle is used lớn balance one element at a time until both sides are equal and the chemical equation is balanced.

1. Count the number of each element on the left and right hand sides

Reactants (Left Hand Side)Products (Right Hand Side)

The number of atoms of each element on both sides of CH3OH + CO = CH3COOH is equal which means that the equation is already balanced and no additional work is needed.

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