The decomposition of n2o5 in carbon tetrachloride proceeds as follows: 2n2o5→4no2+o2 the rate law is first order in n2o5. at 64 ∘c the rate constant is 4.82×10−3s−1. you may want to reference (pages 576 - 581) section 14.3 while completing this problem. part a which of the following is the rate law for this reaction? which of the following is the rate law for this reaction? rate=4.82×10−3s−1[n2o5]2 rate=2.41×10−3s−1[n2o5] rate=4.82×10−3s−1[n2o5] rate=9.64×10−3s−1[n2o5]

r = 2.41 E-3 s-1 [ N2O5 ]

Explanation:

∴ k = 4.82 E-3 s-1

⇒ r =  k [ N2O5 ]/2

⇒ r = (4.82 E-3 s-1 [ N2O5 ])/2

⇒ r = 2.41 E-3 s-1 [ N2O5 ]

rate = 4.82x10⁻³*[N₂O₅]

Explanation:

The rate of a reaction represents how much moles of the reactant is disappearing along the time, and for a reaction with only one reactant, it can be calculated by:

rate = k*[reactant]ⁿ

Where k is the rate constant, [reactant] is the initial concentration of the reactant, and n is the order of the reaction, which is determined by experiments. So, if the rate law is first order, n = 1:

rate = 4.82x10⁻³*[N₂O₅]

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.