The reaction is: Cl2 + 2I- → 2Cl- + I2

The rate of appearance of Cl- is given as d[Cl-]/dt = +0.371 mol L-1 s -1.

According to the stoichiometry of the reaction, 1 molecule of Cl2 reacts to produce 2 molecules of Cl-. Therefore, the rate of disappearance of Cl2 is half the rate of appearance of Cl-.

So, d[Cl2]/dt = d[Cl-]/dt / 2

Substituting the given value:

d[Cl2]/dt = 0.371 mol L-1 s -1 / 2 = 0.1855 mol L-1 s -1

So, the rate of disappearance of Cl2, d[Cl2]/dt, is -0.1855 mol L-1 s -1. The negative sign indicates that Cl2 is being consumed in the reaction.

Question

The reaction is: Cl2 + 2I- → 2Cl- + I2

The rate of appearance of Cl- is given as d[Cl-]/dt = +0.371 mol L-1 s -1.

According to the stoichiometry of the reaction, 1 molecule of Cl2 reacts to produce 2 molecules of Cl-. Therefore, the rate of disappearance of Cl2 is half the rate of appearance of Cl-.

So, d[Cl2]/dt = d[Cl-]/dt / 2

Substituting the given value:

d[Cl2]/dt = 0.371 mol L-1 s -1 / 2 = 0.1855 mol L-1 s -1

So, the rate of disappearance of Cl2, d[Cl2]/dt, is -0.1855 mol L-1 s -1. The negative sign indicates that Cl2 is being consumed in the reaction.

Knowee AI · Accepted Answer

The reaction is: Cl2 + 2I- → 2Cl- + I2

The rate of appearance of Cl- is given as d[Cl-]/dt = +0.371 mol L-1 s -1.

According to the stoichiometry of the reaction, 1 molecule of Cl2 reacts to produce 2 molecules of Cl-. Therefore, the rate of disappearance of Cl2 is half the rate of appearance of Cl-.

So, d[Cl2]/dt = d[Cl-]/dt / 2

Substituting the given value:

d[Cl2]/dt = 0.371 mol L-1 s -1 / 2 = 0.1855 mol L-1 s -1

So, the rate of disappearance of Cl2, d[Cl2]/dt, is -0.1855 mol L-1 s -1. The negative sign indicates that Cl2 is being consumed in the reaction.

For the reaction:Cl2 + 2I- → 2Cl- + I2If d[Cl-]/dt = +0.371 mol L-1 s -1 , what is the value of d[Cl2 ]/dt?

Question

For the reaction:

Solution

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