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.
To determine the value of Δ[I3−]/Δt during the same time interval, we can use the stoichiometric coefficients from the balanced equation.
From the balanced equation, we can see that the stoichiometric 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: Δ[I−]/Δt = 4.8×10−4 M/s
To find Δ[I3−]/Δt, we can use the stoichiometric ratio:
Δ[I3−]/Δt = (1/3) * Δ[I−]/Δt
Δ[I3−]/Δt = (1/3) * 4.8×10−4
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 determine the average rate of consumption of H+ during that time interval, we can use the same approach.
Given: Δ[I−]/Δt = 4.8×10−4 M/s
From the balanced equation, we can see that the stoichiometric coefficient ratio between H+ and I− is 2:3. This means that for every 3 moles of I− consumed, 2 moles of H+ are also consumed.
To find the average rate of consumption of H+, we can use the stoichiometric ratio:
Δ[H+]/Δt = (2/3) * Δ[I−]/Δt
Δ[H+]/Δt = (2/3) * 4.8×10−4
Δ[H+]/Δt = 3.2×10−4 M/s
Therefore, the average rate of consumption of H+ during that time interval is 3.2×10−4 M/s.
To determine the value of Δ[I3-]/Δt, we need to use the balanced equation and compare the stoichiometric coefficients.
The balanced equation for the reaction is:
3I−(aq) + H3AsO4(aq) + 2H+(aq) → I3−(aq) + H3AsO3(aq) + H2O(l)
From the equation, it is clear that the stoichiometric coefficient for I3- is 1. Therefore, the ratio of Δ[I3-]/Δt to Δ[I-]/Δt is 1:3.
Given that Δ[I-]/Δt = 4.8×10^−4 M/s, we can calculate Δ[I3-]/Δt as follows:
Δ[I3-]/Δt = (1/3) * 4.8×10^−4 M/s = 1.6×10^−4 M/s
Thus, the value of Δ[I3-]/Δt during the same time interval is 1.6×10^−4 M/s.
Next, let's determine the average rate of consumption of H+ during that time interval.
From the balanced equation, we can see that each iodide ion reacts with 2 moles of H+. Therefore, the ratio of the rate of consumption of H+ to Δ[I-]/Δt is 2:3.
Given that Δ[I-]/Δt = 4.8×10^−4 M/s, we can calculate the rate of consumption of H+ as follows:
Rate of consumption of H+ = (2/3) * 4.8×10^−4 M/s ≈ 3.2×10^−4 M/s
So, the average rate of consumption of H+ during that time interval is approximately 3.2×10^−4 M/s using two significant figures.