A student conducting the iodine clock experiment accidentally makes an S2O32- stock solution that is too concentrated. How will this affect the rate measurement and the calculated value of k

The question is incomplete the complete question is :

Reaction 1: 3I⁻ (aq) + S₂O₈²⁻ (aq) → I₃⁻ (aq) + 2SO₄²⁻ (aq) slow

Reaction 2: I₃⁻ (aq) + 2S₂O₃²⁻ (aq) → 3I⁻ (aq) + S₄O₆²⁻ (aq) fast

Reaction 3: I₃⁻ (aq) + starch (aq) → 3I⁻ starch (bluish black) fast

A. Not all of the 2S₂O₃²⁻ will react, which will increase the calculated rate.

B. Not all of the 2S₂O₃²⁻ will react, which will decrease the calculated rate.

C. It will take longer for the color to change, as the reaction takes longer when more reactant is present.

D. It will take less time for the color to change, as rate increases with concentration of reactant.

Option D

Explanation:

The reaction time for the experiment will be reduced because the concentrated form of S2O32- reacts very fast with the iodine solution.

The color will change very fast as the reactant is concentrated.Also, the reaction rate(K) of this step would occur faster than the original rate.

This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed.

Here, the concentration has changed.

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2