2S2O3 2- + I2 => S4o6 2- + 2I -

If,in an experiment,0.05 mol S2O3 2- is consumed in 1.0L of solution each second,at what rates are S4O6 2- and I- produced in this solution?

S4O6^2- is 1/2*0.05M and I^- = 0.05M

To determine the rates at which S4O6 2- (thiosulfate) and I- (iodide) are produced in the given chemical reaction, we can use the coefficients of the balanced equation as conversion factors. The balanced equation is:

2S2O3 2- + I2 -> S4O6 2- + 2I -

From the equation, we can see that for every 2 moles of S2O3 2- consumed, 1 mole of S4O6 2- is produced. Similarly, for every 2 moles of S2O3 2- consumed, 2 moles of I- are produced.

Given that the rate of consumption of S2O3 2- is 0.05 mol/s, we can calculate the rates of production of S4O6 2- and I- as follows:

Rate of production of S4O6 2-:
This can be calculated using the following proportion:

(0.05 mol S2O3 2- / 2 mol S2O3 2-) * (1 mol S4O6 2- / 2 mol S2O3 2-) = 0.025 mol/s

Therefore, the rate of production of S4O6 2- in this solution is 0.025 mol/s.

Rate of production of I-:
Using the same approach, we can calculate the rate of production of I- using the following proportion:

(0.05 mol S2O3 2- / 2 mol S2O3 2-) * (2 mol I- / 2 mol S2O3 2-) = 0.05 mol/s

Therefore, the rate of production of I- in this solution is 0.05 mol/s.

In summary, the rate of production of S4O6 2- is 0.025 mol/s, and the rate of production of I- is 0.05 mol/s.