A drop of solution (volume 0.05 mL) contains 3.0× 10-6 moles of H+. If the rate constant of disappearance of H+ is 1.0 × 107 mole litre-1 sec-1. How long would it take for H+ in drop to disappear:

Question

A drop of solution (volume 0.05 mL) contains 3.0× 10-6 moles of H+. If the rate constant of disappearance of H+ is 1.0 × 107 mole litre-1 sec-1. How long would it take for H+ in drop to disappear: (A) 6 × 10-8 sec (B) 6 × 10-7 sec (C) 6 × 10-9 sec (D) 6 × 10-10sec

Tardigrade Admin · Accepted Answer

time Total for drop to disappears(a0-at)=kt at=0 (3.0×10-6/(0.05×10-3)×1.0×107)=t100 % ⇒ t100 %=6×10-9 sec Alternative method Units rate constant =molL-1sec-1⇒ Zero order reaction [A0]-[A]=kT [A0]=(3×10-6×100/0.05×10-3)=(3/50) (3/50)-0=kT ⇒ t=(3/50)×(10-7/1) =6×10-9 sec

Q. A drop of solution (volume 0.05 mL) contains $3.0 \times 1 0^{- 6}$ moles of $H^{+ .}$ If the rate constant of disappearance of $H^{+}$ is $1.0 \times 1 0^{7} m o l e l i t r e^{- 1} se c^{- 1.}$ How long would it take for $H^{+}$ in drop to disappear:

$6 \times 1 0^{- 8}$ sec

$6 \times 1 0^{- 7}$ sec

$6 \times 1 0^{- 9}$ sec

$6 \times 1 0^{- 10}$ sec

Q. A drop of solution (volume 0.05 mL) contains 3.0×10−6 moles of H+. If the rate constant of disappearance of H+ is 1.0×107molelitre−1sec−1. How long would it take for H+ in drop to disappear:

6×10−8 sec

6×10−7 sec

6×10−9 sec

6×10−10sec

Q. A drop of solution (volume 0.05 mL) contains $3.0 \times 1 0^{- 6}$ moles of $H^{+ .}$ If the rate constant of disappearance of $H^{+}$ is $1.0 \times 1 0^{7} m o l e l i t r e^{- 1} se c^{- 1.}$ How long would it take for $H^{+}$ in drop to disappear:

$6 \times 1 0^{- 8}$ sec

$6 \times 1 0^{- 7}$ sec

$6 \times 1 0^{- 9}$ sec

$6 \times 1 0^{- 10}$ sec