Consider the following reaction: 2H2O2(aq)→2H2O(l)+O2(g) Use the graph to calculate the following.?
The instantaneous rate of formation of O2 at 50s.
2. If the initial volume of the H2O2 solution is 1.6 L , what total amount of O2 (in moles) is formed in the first 50 s of reaction?
This is the best answer for Consider the following reaction: 2H2O2(aq)→2H2O(l)+O2(g) Use the graph to calculate the following.?
for the general reaction
.. . A –> B + C
rate can be defined as
.. rate = – d[A] / dt
for this reaction..
.. 2 H2O2 —> 2 H2O + 1 O2
rate of reaction = -d[H2O2] / dt = -slope of the tangent to any point on the curve [H2O2] vs t
so.. do this
.. (1) draw a tangent to the curve at t=50,
.. (2) pick any 2 data points on that tangent line and calc the slope
.. .. . (Y2 – Y1) / (X2 – X1)
.. (3) the rate of the rxn… with respect to H2O2 consumption = – slope of that tangent line
.. (4) from the balanced equation, 2 H2O2 –> 1 O2
.. …..so the rate of O2 production = 1/2 the rate of H2O2 consumption.
rather than guess at the tangent line, I actually, read the [H2O2] concentrations for times 0, 10, 20, 30, etc and plotted time vs ln[H2O2]…. which turned out to be linear with
.. y = -0.029424x – 0.008231…. ..r^2 = 0.999726
so that I could calculate the [H2O2] value at say t = 48sec and t = 52 sec and use those to generate the slope of the tangent line… looked something like this
.. . .t.. .. . ..ln[H2O2].. .. [H2O2]
.. . .48.. .. .-1.42538.. .0.240416
.. . .52.. .. .-1.54348.. .0.213637
and slope = (0.240416 – 0.213637) / (48 – 52) = -0.006695 M H2O2 / sec
so that rate of reaction (in terms of H2O2 consumption = +0.006695 M / sec
and rate of generation of O2 = 0.5x+0.006695 M (H2O2) / sec = 3.35×10^-3 M/sec
i.e… 3.35×10^-3 moles O2 / (L x sec)
At t = 50 sec
.. [H2O2] = 0.225M
so from t = 0 sec to t = 50sec, the concentration of H2O2 has decreased from 1.0M to 0.225M
If we stated with 1.6L solution, we started with
.. 1.6L x (1mol H2O2 / 1L) = 1.6mol H2O2
now IF…. IF.. all that H2O2 decomposed, we would have generated
.. 1.6mol H2O
and that would add this much water to our solution
.. 1.6mol x (18g / mol) x (0.001L / g) = 0.03L of H2O
in other words.. IF all the H2O2 was consumed, our amount of water present would add to the volume of water we started with but NOT significantly.. So… this is where I going with this.
.. “we can ignore the volume of water produced by the rxn”
that makes things MUCH easier
.. initial moles H2O2 = 1.6mol H2O2
.. final moles H2O2 = 1.6L x (0.225mol / L) = 0.36mol H2O2
.. moles H2O2 consumed = 1.6 – 0.36 = 1.24mol H2O2
and since 2 H2O2 –> 1 O2
.. moles O2 produced = 0.5 x 1.24 = 0.62 mol O2 produced… (2 sig fig)