For the reaction:
2A + B → A2B
the rate = k[A][B]2with k= 2.0 x 10-6mol-2L2s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1.
The initial rate of the reaction is
Rate = k [A][B]2
= (2.0 × 10 - 6mol - 2L2s - 1) (0.1 mol L - 1) (0.2 mol L - 1)2
= 8.0 × 10 - 9mol - 2L2s - 1
When [A] is reduced from 0.1 mol L - 1to 0.06 mol - 1, the concentration of A reacted = (0.1 - 0.06) mol L - 1 = 0.04 mol L - 1
Therefore, concentration of B reacted= 1/2 x 0.04 mol L-1 = 0.02 mol L - 1
Then, concentration of B available, [B] = (0.2 - 0.02) mol L - 1
= 0.18 mol L - 1
After [A] is reduced to 0.06 mol L - 1, the rate of the reaction is given by,
Rate = k [A][B]2
= (2.0 × 10 - 6mol - 2L2s - 1) (0.06 mol L - 1) (0.18 mol L - 1)2
= 3.89 mol L - 1s - 1
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(ii) Potassium tetracyanonickelate(II)
(iii) Tris(ethane-1,2-diamine) chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate
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(ii) Write IUPAC names of all the isomers.
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(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec-butyl ketone
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(iii) 4-tert. Butyl-3-iodoheptane
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(iii) Semicarbazone
(iv) Aldol
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(vii) Ketal
(vii) Imine
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How come we halved the concentration of A to find the new one of B
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Second part of this answer\'s method is correct but answer is wrong...it will be = 3.888*10^-9 mol-2L2s-1