Experimental data have shown that the rate law for the reaction 2 hgcl2 (aq) + c2o4 2 - (aq) → 2 cl - (aq) + 2 co2 (g) + hg2cl2 (s) is: rate = k[hgcl2][c2o4 2 - ]2 how will the rate of reaction change if the concentration of c2o4 2 - is tripled and the concentration of hgcl2 is doubled?

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Chemistry

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10 July, 10:10

Experimental data have shown that the rate law for the reaction 2 hgcl2 (aq) + c2o4 2 - (aq) → 2 cl - (aq) + 2 co2 (g) + hg2cl2 (s) is: rate = k[hgcl2][c2o4 2 - ]2 how will the rate of reaction change if the concentration of c2o4 2 - is tripled and the concentration of hgcl2 is doubled?

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Kayley Delgado · Answer 1

When the rate = K [ HgCl2] [C2O4 2-]^2

if the concentration of [ C2O4 2 - ] is tripled so we will substitute by [3 (C2O4 2-) ] instead of [ C2O4 2 - ]

and if the concentration of [HgCl2] is doubled so we will substitute by [2 (HgCl2) ] instead of [HgCl2]

So the new rate will be = K [2 (Hgcl2) ] [3 (C2O4 2-) ]^2