Determine the rate of consumption of I− if the initial concentration of IO3− was 0.6634 M and decreased to 0.5745 M after 0.3964 hours. Use the reaction below.
IO3− + 5I− + 6H+ ↔ 3I2 + 3H2O
0.6634-0.5745 = 0.0889 M in 0.3964 hrs = about 0.22 M/hr (but you may want to change that to M/s.
Then rate change IO3^- x (5 mol I/1 mol IO3) = ? rate change for I^-
To determine the rate of consumption of I-, we need to use the equation for the rate of reaction.
First, we need to determine the change in concentration of I-. The initial concentration of IO3− is given as 0.6634 M, and the final concentration is given as 0.5745 M. The change in concentration is the difference between these two values:
Change in concentration of I- = Final concentration of I- - Initial concentration of I-
= 0.5745 M - 0.6634 M
= -0.0889 M
Next, we need to determine the change in time. The time is given as 0.3964 hours.
Now, we can use the rate equation to calculate the rate of consumption of I-. The rate equation is given by:
Rate = Change in concentration of I- / Change in time
Plugging in the values we calculated:
Rate = -0.0889 M / 0.3964 hours
To find the units of rate, we divide the units of concentration by the units of time:
Rate = -0.0889 M / 0.3964 hours
= -0.224 M/h
Therefore, the rate of consumption of I- is -0.224 M/h (negative sign indicates consumption).