My Courses Homepage - CHM-122-002 Pearson MyLab and Mastering Course Home Master

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My Courses Homepage - CHM-122-002 Pearson MyLab and Mastering Course Home MasteringChemistry: Chapter... Chegg Study | Guided S ID Sessio Homepago- CHM-1 For More Practice 18.8 Enhanced with Feedback previous 1 of 15 | next » For More Practice 18.8- Enhanced with Feedback Part A Consider the following reaction: Calculate Gim again at 25°C, but use the equation "n,-Arn-TASrxn to determine the value Express the free energy to four significant figures and include the appropriate units. 2C0(g) + 2N0(g) 2CO2 (g)--N2(g) when the standard G2 values are used to calculate the value of Gron at 25°C, we get-689.6 kJ Standard thermodynamic quantities for selected substances at 25 Substance | , (kJ/mol) | Gi (kJ/mol) Sa (J/(mol . K)) 011 ? Grxn = Value Units 02(g) 03(g) CO(g) CO2(g) N2(g) NO(g) NO2(g) 205.2 238.9 197.7 213.8 191.6 210.8 240.1 Submit My Answers Give Up 142.7 110.5 -393.5 163.2 -137.2 -394.4 Part B This question will be shown after you complete previous question(s). 91.3 87.6 33.2 51.3 Part C This question will be shown after you complete previous question(s). You may want to reference ( completing this problem. -pagesB64-869) Section 18.8 while Part D This question will be shown after you complete previous question(s). Continue

Explanation / Answer

A)

we have:

Hof(CO(g)) = -110.5 KJ/mol

Hof(NO(g)) = 91.3 KJ/mol

Hof(CO2(g)) = -393.5 KJ/mol

Hof(N2(g)) = 0.0 KJ/mol

we have the Balanced chemical equation as:

2 CO(g) + 2 NO(g) ---> 2 CO2(g) + N2(g)

deltaHo rxn = 2*Hof(CO2(g)) + 1*Hof(N2(g)) - 2*Hof( CO(g)) - 2*Hof(NO(g))

deltaHo rxn = 2*(-393.5) + 1*(0.0) - 2*(-110.5) - 2*(91.3)

deltaHo rxn = -748.6 KJ

we have:

Sof(CO(g)) = 197.7 J/mol.K

Sof(NO(g)) = 210.8 J/mol.K

Sof(CO2(g)) = 213.8 J/mol.K

Sof(N2(g)) = 191.6 J/mol.K

we have the Balanced chemical equation as:

2 CO(g) + 2 NO(g) ---> 2 CO2(g) + N2(g)

deltaSo rxn = 2*Sof(CO2(g)) + 1*Sof(N2(g)) - 2*Sof( CO(g)) - 2*Sof(NO(g))

deltaSo rxn = 2*(213.8) + 1*(191.6) - 2*(197.7) - 2*(210.8)

deltaSo rxn = -197.8 J/K

deltaHo = -748.6 KJ/mol

deltaSo = -197.8 J/mol.K

= -0.1978 KJ/mol.K

T= 25.0 oC

= (25.0+273) K

= 298 K

we have below equation to be used:

deltaGo = deltaHo - T*deltaSo

deltaGo = -748.6 - 298.0 * -0.1978

deltaGo = -689.6556 KJ

Answer: -689.7 KJ

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