For the irreversible process, A + B → products, the rate is first-order w.r.t. A and second-order w.r.t. B. If 1.0 mol each of A and B introduced into a 1.0 L vessel, and the initial rate was 1.0 × 10-2 mol L-1 s-1 rate when half reactants have been turned into products is:

Question

For the irreversible process, A + B → products, the rate is first-order w.r.t. A and second-order w.r.t. B. If 1.0 mol each of A and B introduced into a 1.0 L vessel, and the initial rate was 1.0 × 10-2 mol L-1 s-1 rate when half reactants have been turned into products is: (A) 1.25 × 10-3 mol L-1 s- (B) 1.0 × 10-3 mol L-1 s- (C) 2.50 × 10-3 mol L-1 s- (D) 2.0 × 10-3 mol L-1 s-

Tardigrade Admin · Accepted Answer

A+B→ product r=K[A]1[B]2 r1=K[1]1[1]1=1×10-2 (K=1×10-2) r1=K[(1/2)][(1/2)]2=1×10-2×(1/8) =1.25×10-3mol L-1S-1

$1.25 \times 1 0^{- 3} m o l L^{- 1} s^{-}$

$1.0 \times 1 0^{- 3} m o l L^{- 1} s^{-}$

$2.50 \times 1 0^{- 3} m o l L^{- 1} s^{-}$

$2.0 \times 1 0^{- 3} m o l L^{- 1} s^{-}$

$A + B \to$ product
$r = K [A]^{1} [B]^{2}$
$r_{1} = K [1]^{1} [1]^{1} = 1 \times 1 0^{- 2} (K = 1 \times 1 0^{- 2})$
$r_{1} = K [\frac{1}{2}] [\frac{1}{2}]^{2} = 1 \times 1 0^{- 2} \times \frac{1}{8}$
$= 1.25 \times 1 0^{- 3} m o l L^{- 1} S^{- 1}$

Q. For the irreversible process, A+B→ products, the rate is first-order w.r.t. A and second-order w.r.t. B. If 1.0mol each of A and B introduced into a 1.0L vessel, and the initial rate was 1.0×10−2molL−1s−1 rate when half reactants have been turned into products is:

1.25×10−3molL−1s−

1.0×10−3molL−1s−

2.50×10−3molL−1s−

2.0×10−3molL−1s−

A+B→ product r=K[A]1[B]2 r1​=K[1]1[1]1=1×10−2(K=1×10−2) r1​=K[21​][21​]2=1×10−2×81​ =1.25×10−3molL−1S−1

$1.25 \times 1 0^{- 3} m o l L^{- 1} s^{-}$

$1.0 \times 1 0^{- 3} m o l L^{- 1} s^{-}$

$2.50 \times 1 0^{- 3} m o l L^{- 1} s^{-}$

$2.0 \times 1 0^{- 3} m o l L^{- 1} s^{-}$

$A + B \to$ product
$r = K [A]^{1} [B]^{2}$
$r_{1} = K [1]^{1} [1]^{1} = 1 \times 1 0^{- 2} (K = 1 \times 1 0^{- 2})$
$r_{1} = K [\frac{1}{2}] [\frac{1}{2}]^{2} = 1 \times 1 0^{- 2} \times \frac{1}{8}$
$= 1.25 \times 1 0^{- 3} m o l L^{- 1} S^{- 1}$