What is the oxidation number for chromium in each of these compounds. show your work. a. k2cro4 b. cr2o3

A. + 6.

B. + 3.

Explanation:

A. K₂CrO₄:

We have that the sum of the oxidation no. of different elements in the compound multiplied by its no. is equal to the overall charge of the compound.

∵ 2(oxidation no. of K) + (oxidation no. of Cr) + 4(oxidation no. of O) = 0.

oxidation no. of K = + 1, oxidation no. of O = - 2.

∴ 2(+ 1) + (oxidation no. of Cr) +4(- 2) = 0.

∴ 2 + (oxidation no. of Cr) - 8 = 0.

∴ (oxidation no. of Cr) - 6 = 0.

∴ oxidation no. of Cr = + 6.

B. Cr₂O₃:

∵ 2(oxidation no. of Cr) + 3(oxidation no. of O) = 0.

oxidation no. of O = - 2.

∴ 2(oxidation no. of Cr) + 3(- 2) = 0.

∴ 2(oxidation no. of Cr) - 6 = 0.

∴ 2(oxidation no. of Cr) = + 6.

∴ (oxidation no. of Cr) = + 6/2 = + 3.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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explanation:

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