Consider the following equilibrium:2NO2g N2O4g= ΔG0−5.4kJNow suppose a reaction vessel is filled with 8.33atm of dinitrogen tetroxide N2O4 at 294.°C. Answer the following questions about this system:Under these conditions, will the pressure of N2O4 tend to rise or fall? rise fall Is it possible to reverse this tendency by adding NO2?In other words, if you said the pressure of N2O4 will tend to rise, can that be changed to a tendency to fall by adding NO2? Similarly, if you said the pressure of N2O4 will tend to fall, can that be changed to a tendency to rise by adding NO2? yes no If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO2 needed to reverse it.Round your answer to 2 significant digits.atm

The spontaneous chemical reaction proceeds on its own in the forward direction when the change in the free energy (∆G) value is negative. The relationship between ∆G and the reaction quotient (Q) can be furnished as follows: ∆G=∆Go+RTln Q ...(1)∆Go=Standard free energy changeR=Gas constantT=Temperature of the system In an equilibrium reaction, the rate of the conversion of the reactant to products is observed to be equal to the opposite conversion, i.e., the conversion of products to reactants at a specified temperature. The magnitude of the equilibrium constant is furnished as the ratio of the concentration of the products to the concentration of the reactants.i) Pressure change: When the equilibrium reverses its direction, the magnitude of ∆G becomes zero. Thus, equation 1 is changed as follows: ∆Go=-RT ln Q ...(2) For the conversion of NO2 to N2O4, the expression for Q is furnished as follows: Q=pN2O4pNO22 Substitute the parameters in equation 2 to determine pNO2 as follows: ∆Go=-5.4 kJR=8.314 J/K mol=1 kJ1000 J×8.314 J/K mol=0.008314 kJ/K molT=294°C=(294+273)K=567 K-5.4 kJ=-(0.008314 kJ/K mol)(567 K)ln 8.33 atmpNO221.145=ln 8.33-2 ln pNO21.145-2.119=-2 ln pNO2-0.974=-2 ln pNO20.487= ln pNO2 pNO2=1.62 atm Since the final pressure is decreased from 8.33 atm to 1.62 atm, the answer is "fall."ii) The equilibrium reaction: 2NO2(g)↔N2O4(g) From the reaction, it is observed that the addition of nitrogen dioxide will favor the formation of N2O4. Hence, the reaction an be reversed by the addition of NO2. Thus, the answer is "yes." iii) Determination of pressure: -5.4 kJ=-(0.008314 kJ/K mol)(567 K)lnpN2O4pNO221.145=lnpN2O4pNO22pN2O4pNO22=3.14pN2O4=3.1×2×pNO2=6.2×pNO2 Hence, the pressure of N2O4 should be 6.28 times the pressure of NO2.

Answered: Consider the following equilibrium:…

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

Consider the following equilibrium:

2NO2g

N2O4g

= ΔG0−5.4kJ

Now suppose a reaction vessel is filled with

8.33atm

of dinitrogen tetroxide

N2O4

294.°C

. Answer the following questions about this system:

Under these conditions, will the pressure of N2O4 tend to rise or fall?	rise
fall
Is it possible to reverse this tendency by adding NO2 ? In other words, if you said the pressure of N2O4 will tend to rise, can that be changed to a tendency to fall by adding NO2 ? Similarly, if you said the pressure of N2O4 will tend to fall, can that be changed to a tendency to rise by adding NO2 ?	yes
no
If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO2 needed to reverse it. Round your answer to 2 significant digits.	atm

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