For the following balanced equation: 2 NO(g) + O_2(g) rightarrow N_2O_4(g) Calcu

Question

For the following balanced equation: 2 NO(g) + O_2(g) rightarrow N_2O_4(g) Calculate Delta H degree and Delta S degree from values in Table 6 and use these values to calculate Delta G degree at 800 K. a. -75.2 kJ/mol b. -77.4 kJ/mol c. -241.4 kJ/mol d. +8.2 kJ/mol e. +86.6 kJ/mol Consider the following correctly balanced redox reaction in acidic solution: 3 Sn^4+(aq) + 2 Au(s) rightarrow 3 Sn^2+(aq) + 2Au^3+(aq) (acidic) It can be calculated that E_cell degree = - 135 V. What is E_cell when [Sn^4+] = 4.2 times 10^-6 M. [Sn^2+] = 0.15 M, and [Au^3+] = 1.7 times 10^-3 M? a. -1.27 V b. -1.53 V c. -1.83 V d. -1.43 V e. +1.27 V

Explanation / Answer

1 .N2O4----> 2NO2   deltaH0= 58 KJ (1) and NO+0.5O2---->NO2   DeltaH0 = -56 KJ (2)

Eq.2*2 gives 2NO+O2---->2NO2    deltaH0= -112 KJ (2A)

reversing Eq.1 gives 2NO2---------->N2O4 deltaH= -58 Kj (1A)

Addition of 1A and 2A equations gives 2NO+O2---->N2O4 deltaH= -112-58= -170 Kj

Standard entropy change data : N2O4 =304.3 J/Kmol, NO :210.7            O2 : 205.1J/Kmole

Standard entropy change of reaction = sum of entropy of products- sum of entropy of reactants =

304.3- (2*210.7 +205.1)=-322.2 J/K (1, 2 and 1 are coefficients of N2O4, O2 and NO)

deltaG at 800 K= deltaH- TdeltaS= -170 + 800*322.2/1000 Kj =87.76 Kj/mole ( close to 86.6 Kjmole e is correct)

2. E= EO- 0.0591/n * log Q

n =number of eletrons exchanged

Au+3 +3e- -------> Au   (1) reversin it gives Au ----------->Au+3 +3e- (1A) and Sn+4 +2e------>Sn+2 (2),

multiply Eq.1 with 2 and Eq.2 with 3 and additino gives required equation where there is exchange of 6 electrons.

n= 6

Q= [Au+3] 2 [Sn+2] 3/ [Sn+4]3 = (1.7*10-3)2* (0.15)3/ [ 4.2*10-6)3 =1.32*108

logQ = 8.119

E= -1.35- (0.0591/6)*8.119=-1.43 V ( d is correct)

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---