Concentration and time data for a given reaction at a certain temperature are sh

Question

Concentration and time data for a given reaction at a certain temperature are shown in Option IV. Write the balanced equation for the reaction, and place the concentration and time data in the appropriate columns in the table below Then, calculate ln [X] 1/[X] for each [X] value and enter these values in the table. Balanced Equation: 2N_2O_5 Rightarrow 4NO_2 + O_2 Plot a graph of [X] vs. time Plot a graph of ln [X] vs. time Plot a graph of 1/[X] vs. time Examine your three plots. Which one appears to be the most linear? Using the linear plot, determine the value of the rate constant for the reaction. Write the rate law for the reaction using the rate constant determined above. Write the integrated form of the rate law and use this equation to answer the following questions a) Determine the concentration of N_2O_5 at 15 min b) At what time would the concentration of NO_2 be 25 min?

Explanation / Answer

for any order reaction, -dCA/dt= KCAn, n is order of reaction.

n=0 refer to zero order reaction. hence -dCA/dt= K, when integrated, the equation becomes

CA= CAO-Kt, where CA= Concentration of A at any time, CAO= initial concentration, K is the rate constant

hence a plot of CA vs t gives a straight line whose slope is -K.

for 1st order reaction, -dCA/dt= KCA, which when integrated gives lnCA= lnCAO-Kt. so for 1st order reaction, the plot of lnCA vs t gives strainght line whose slope is -K.

for second order reaction, -dCA/dt= KCA2, which when integrated gives 1/CA =1/CAO+Kt, so a plot of 1/CA vs t gives straight line whose slope is K.

all the three pots for the given data are drawn and shown below.

the highest value of R2 value will be the best fit. Accordingly, 1st order equation having highest R2 value 0.999 will be the best fit.

hence ln CA= -0.033*t-.3.221

slope K =0.033/min

at 15 min, lnCA= -0.033*15-3.221, CA =0.024 M

at 25 min, lnCA= -0.033*25-3.221, CA=0.0175

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