Question

If you collect 5.74 mL of O 2 at 298 K and 1.00 atm over 60.0 seconds from a reaction solution of 5.08 mL, what is the initial rate of the reaction

Answer 1

Answer:

7.71 × 10⁻⁴ M/s

Explanation:

The initial rate of the reaction can be expressed by using the formula:

$\dfrac{\Delta [O_2]}{\Delta t}$

where the number of moles of O₂ = $\dfrac{PV}{RT}$

where;

Pressue P = 1.00 atm

Volume V =5.74mL =  (5.74 /1000) L

Rate R = 0.082 L atm/mol.K

Temperature = 298 K

$= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}$

= 2.35 × 10⁻⁴ mol

Δ[O₂] = $\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }$

Δ[O₂] = $\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}$

Δ[O₂]  = 0.04626 M

The initial rate = $\dfrac{\Delta [O_2]}{\Delta t}$

= $\dfrac{0.04626}{60}$

= 7.71 × 10⁻⁴ M/s