The oxidation of iodide ion by arsenic acid, H3AsO4, is described by the balanced equation
3I−(aq)+H3AsO4(aq)+2H+(aq)→ I3−(aq)+H3AsO3(aq)+H2O(l)
If −Δ[I−]/Δt = 4.8×10−4M/s , what is the value of Δ[I3−]/Δt during the same time interval?
Express your answer using two significant figures.
What is the average rate of consumption of H+ during that time interval?
Express your answer using two significant figures.
From the balanced equation, we can see that the ratio of the coefficient of I3- to I- is 1:3. This means that for every 1 mole of I3- produced, 3 moles of I- are consumed.
Therefore, the rate of change of [I3-] is three times the rate of change of [I-].
Given that Δ[I-]/Δt = 4.8 × 10^-4 M/s, we can find Δ[I3-]/Δt by multiplying 4.8 × 10^-4 M/s by 3.
Δ[I3-]/Δt = (4.8 × 10^-4 M/s) x 3 = 1.44 × 10^-3 M/s
So, the value of Δ[I3-]/Δt during the same time interval is 1.44 × 10^-3 M/s.
To find the average rate of consumption of H+ during the time interval, we look at the coefficient of H+ in the balanced equation, which is 2.
Therefore, the rate of change of [H+] is two times the rate of change of [I-].
Given that Δ[I-]/Δt = 4.8 × 10^-4 M/s, we can find Δ[H+]/Δt by multiplying 4.8 × 10^-4 M/s by 2.
Δ[H+]/Δt = (4.8 × 10^-4 M/s) x 2 = 9.6 × 10^-4 M/s
So, the average rate of consumption of H+ during that time interval is 9.6 × 10^-4 M/s.
To find the value of Δ[I3−]/Δt, we can use the stoichiometric coefficients from the balanced equation. In the balanced equation, we can see that the coefficient ratio between I− and I3− is 3:1. This means that for every 3 moles of I− consumed, 1 mole of I3− is produced.
Given that −Δ[I−]/Δt = 4.8×10−4 M/s, we can substitute this value into the stoichiometry to find the value of Δ[I3−]/Δt.
Δ[I3−]/Δt = (Δ[I−]/Δt) × (1/3)
Δ[I3−]/Δt = (4.8×10−4 M/s) × (1/3)
Δ[I3−]/Δt = 1.6×10−4 M/s
Therefore, the value of Δ[I3−]/Δt during the same time interval is 1.6×10−4 M/s.
To find the average rate of consumption of H+ during that time interval, we can use the stoichiometric coefficient of H+ in the balanced equation, which is 2.
The average rate of consumption of H+ = 2 × (Δ[I−]/Δt)
Average rate of consumption of H+ = 2 × (4.8×10−4 M/s)
Average rate of consumption of H+ = 9.6×10−4 M/s
Therefore, the average rate of consumption of H+ during that time interval is 9.6×10−4 M/s.