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i) Based on the rate and concentration given in the table above determine the order of reaction for each reactant (without using graph paper). ii) Write the rate equation for this reaction. iii) Calculate value of k using the rate equation.

i) Based on the rate and concentration given in the table above determine the order of reaction for each reactant (without using graph paper). ii) Write the rate equation for this reaction. iii) Calculate value of k using the rate equation.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question i,ii,iii

For the reaction: 2NO + Cl2 → 2 NOCI at 300K, the following data was obtained. Initial Concentration Initial Rates Exp. No. [NO] 0.01 0.01 0.02 [Cl2] 0.01 0.02 (moldm³s1) 1.2 x 10-4 2.4 x 104 1 2 0.02 9.6 x 104
Transcribed Image Text:For the reaction: 2NO + Cl2 → 2 NOCI at 300K, the following data was obtained. Initial Concentration Initial Rates Exp. No. [NO] 0.01 0.01 0.02 [Cl2] 0.01 0.02 (moldm³s1) 1.2 x 10-4 2.4 x 104 1 2 0.02 9.6 x 104
i) Based on the rate and concentration given in the table above determine the order of reaction for each reactant (without using graph paper). ii) Write the rate equation for this reaction. iii) Calculate value of k using the rate equation. iv) Determine the half time for nitrogen monoxide. v) Calculate the rate when concentration of both reactants is 0.025 mol/L.
Transcribed Image Text:i) Based on the rate and concentration given in the table above determine the order of reaction for each reactant (without using graph paper). ii) Write the rate equation for this reaction. iii) Calculate value of k using the rate equation. iv) Determine the half time for nitrogen monoxide. v) Calculate the rate when concentration of both reactants is 0.025 mol/L.
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