Chapter 12, Problem 42E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# At 500 K in the presence of a copper surface, ethanol decomposes according to the equation C 2 H 5 OH ( g ) → CH 3 CHO ( g ) + H 2 ( g ) The pressure of C2H5OH was measured as a function of time and the following data were obtained: Time(s) P C 2 H 5 OH ( torr ) 0 250. 100. 237 200. 224 300. 211 400. 198 500. 185 Since the pressure of a gas is directly proportional to the concentration of gas, we can express the rate law for a gaseous reaction in terms of partial pressures. Using the above data, deduce the rate law, the integrated rate law, and the value of the rate constant, all in terms of pressure units in atm and time in seconds. Predict the pressure of C2H5OH after 900. s from the start of the reaction. (Hint: To determine the order of the reaction with respect to C2H5OH, compare how the pressure of C2H5OH decreases with each time listing.)

Interpretation Introduction

Interpretation: The table of pressure of ethanol (C2H5OH) as a function of time is given. Therate law, the integrated rate law and the value of rate constant is to be calculated based on these data. Also the pressure of (C2H5OH) after 900s is to be predicted.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

To determine: The rate law, the integrated rate law and the value of rate constant for the given reaction; the pressure of (C2H5OH) after 9.00s .

Explanation

The given reaction is,

C2H5OH(g)C2H4(g)+H2(g)

The table of pressure as a function of time is,

 Time (s) PC2H5OH(torr) 0 250 100 237 200 224 300 211 400 198 500 185

The graph is drawn from the above values of pressure and time.

Figure 1

The plot of [A] versus time is straight line with negativeslope. It means that the graph is plotted for the zero order reaction.

The rate law gives the relation between reaction rate and concentration of reactants. The rate law is represented as,

Rate=k[A]a

C2H2OH

Where,

• k is rate constant.
• [A] is concentration of reactant.
• a is reaction order (zero for zero order reaction).

Substitute the value of a in above equation.

Rate=k[PC2H5OH]0=k(1)=k

To determine: The integrated rate law for the given reaction.

Given

The given reaction is,

C2H5OH(g)C2H4(g)+H2(g)

The rate law in integral form is known as integrated rate law. It gives the relation between concentration of reactant or product with time. The integral of rate law is represented as,

[C2H5OH]t=kt+[PC2H5OH]0

Where,

• k is rate constant.
• t is time.
• [C2H5OH]t isconcentration of reactant at time t .

To determine: The value of rate constant for the given reaction.

The initial time is 0s .

The final time is 100s .

The initial pressureis 250torr .

The finalpressureis 237torr .

The conversion of (torr) into (atm) is done as,

1torr=1760atm

Hence, the conversion of 250torr into (atm) is,

250torr=(250×1760)atm=(250760)atm

The conversion of 237torr into (atm) is,

237torr=(237×1760)atm=(237760)atm

Formula

The slope of graph is calculated using the formula,

Slope=ΔyΔx=y2y1x2x1

Where,

• y2 is final pressure

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