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In a particular experiment, it was found that when O₂(g) and CO(g) were mixed and allowed to react according to the equation 2CO(g) + O₂(g) → 2CO₂(g) the O₂ concentration had decreased by 0.041 mol L-1 when the reaction reached equilibrium. How had the concentrations of CO and CO₂ changed? [CO] decrease = i [CO₂] increase = i mol/L mol/L

In a particular experiment, it was found that when O₂(g) and CO(g) were mixed and allowed to react according to the equation 2CO(g) + O₂(g) → 2CO₂(g) the O₂ concentration had decreased by 0.041 mol L-1 when the reaction reached equilibrium. How had the concentrations of CO and CO₂ changed? [CO] decrease = i [CO₂] increase = i mol/L mol/L

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
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In a particular experiment, it was found that when O₂(g) and CO(g) were mixed and allowed to react according to the equation: \[ 2\text{CO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{CO}_2(g) \] the O₂ concentration had decreased by 0.041 mol L⁻¹ when the reaction reached equilibrium. How had the concentrations of CO and CO₂ changed? - [CO] decrease = [input] mol/L - [CO₂] increase = [input] mol/L
Transcribed Image Text:In a particular experiment, it was found that when O₂(g) and CO(g) were mixed and allowed to react according to the equation: \[ 2\text{CO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{CO}_2(g) \] the O₂ concentration had decreased by 0.041 mol L⁻¹ when the reaction reached equilibrium. How had the concentrations of CO and CO₂ changed? - [CO] decrease = [input] mol/L - [CO₂] increase = [input] mol/L
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To find the concentration decrease , first we need to write relation between rate of decrease of reactants and rate of increase of products for given reaction. Then. We can use the decrease in concentration given for O2 . 

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