Question: The reaction [2 mathrm{NO}(g)+mathrm{Cl}_{2}(g) longrightarrow 2 mathrm{NOCl}(g)] was studied at (-10^{circ} mathrm{C}). The following results were obtained, where [text { Rate }=-frac{dleft[mathrm{Cl}_{2}ight]}{d t}] a. What
The reaction
\[2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\]
was studied at \(-10^{\circ} \mathrm{C}\). The following results were obtained, where
\[\text { Rate }=-\frac{d\left[\mathrm{Cl}_{2}ight]}{d t}\]
![[NO]o (mol/L) 0.10 0.10 0.20 [Cl]o (mol/L) 0.10 0.20 0.20 Initial Rate](https://dsd5zvtm8ll6.cloudfront.net/images/question_images/1706/1/7/3/26765b22353a89921706173267454.jpg)
a. What is the rate law?
b. What is the value of the rate constant?
[NO]o (mol/L) 0.10 0.10 0.20 [Cl]o (mol/L) 0.10 0.20 0.20 Initial Rate (mol L min) 0.18 0.36 1.45
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To determine the rate law for the given reaction we need to analyze how changes in the concentrations of the reactants affect the rate of the reaction The rate law can be written in the general form Rate k NOx Cl2y where k is the rate constant NO and Cl2 are the concentrations of the reactants and x and y are the orders of the reaction with respect to NO and Cl2 respectively We will use the provided initial rates and concentrations to deduce the values of x and y Lets examine how the initial rate changes when the concentration ... View full answer
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