The reaction RBr + Nu- -> RNu + Br- was studied at constant temperature. Initial rates were: Run 1, [RBr] = 0.10 M, [Nu-] = 0.10 M, rate = 4.8 x 10^-5 M s^-1; Run 2, 0.20 M, 0.10 M, 9.6 x 10^-5 M s^-1; Run 3, 0.10 M, 0.30 M, 1.44 x 10^-4 M s^-1. Select the correct statements.

Doubling [Nu-] while halving [RBr] leaves the initial rate unchanged.

The reaction is first order in RBr and first order in Nu-.

Doubling [Nu-] while halving [RBr] leaves the initial rate unchanged.

The reaction is first order in RBr and first order in Nu-.

At [RBr] = 0.05 M and [Nu-] = 0.40 M the initial rate is 1.2 x 10^-4 M s^-1.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3o3092

The reaction $RBr + Nu X^{-} RNu + Br X^{-}$ was studied at constant temperature. Initial rates were: Run 1, $[RBr] = 0.10 M$ , $[Nu X^{-}] = 0.10 M$ , rate $= 4.8 \times 1 0^{- 5} M s^{- 1}$ ; Run 2, $0.20 M$ , $0.10 M$ , $9.6 \times 1 0^{- 5} M s^{- 1}$ ; Run 3, $0.10 M$ , $0.30 M$ , $1.44 \times 1 0^{- 4} M s^{- 1}$ . Select the correct statements.

ADoubling $[Nu X^{-}]$ while halving $[RBr]$ leaves the initial rate unchanged.
BThe reaction is first order in $RBr$ and first order in $Nu X^{-}$ .
CAt $[RBr] = 0.05 M$ and $[Nu X^{-}] = 0.40 M$ the initial rate is $1.2 \times 1 0^{- 4} M s^{- 1}$ .
D $k = 4.8 \times 1 0^{- 3} L mo l^{- 1} s^{- 1}$ .

Run 2 doubles

[RBr]

and doubles the rate; run 3 triples

[Nu X^{-}]

and triples the rate, so the rate law is rate

= k [RBr] [Nu X^{-}]

. From run 1,

k = 4.8 \times 1 0^{- 5} / (0.10 \times 0.10) = 4.8 \times 1 0^{- 3} L mo l^{- 1} s^{- 1}

. Doubling one concentration and halving the other leaves the product, and the rate, unchanged. At 0.05 M and 0.40 M the rate is

4.8 \times 1 0^{- 3} \times 0.05 \times 0.40 = 9.6 \times 1 0^{- 5} M s^{- 1}

, not

1.2 \times 1 0^{- 4}