The following standard free energy changes are given for 25 degree C. N_2(g)+3H_
ID: 519315 • Letter: T
Question
The following standard free energy changes are given for 25 degree C. N_2(g)+3H_2(g) 2NH_3(g) Delta G degree = -33.0 kJ 4NH_3(g)+5 O_2(g) 4NO(g) + 6H_2O(1) Delta G degree = -1011. kJ N_2(g)+O_2(g) 2NO (g) Delta G degree = +173.1 kJ N_2(g)+2O_2(g) 2NO_2 (g) Delta G degree = +102.6 kJ 2N_2(g)+O_2(g) 2N_2O (g) Delta G degree = +208.4 kJ Combine the above equations, as necessary, to obtain Delta G^degree values for the following reactions (review Hess's Law) N_2O(g) + 3/2O_2 (g) 2 NO_2 (g) Delta G^degree = ? 2H_2(g)+ O_2(g) 2H_2O(l) Delta G^degree = ? 2NH_3(g) + 2O_2 (g) N_2O (g) + 3H_2O(1) Delta G^degree = ? Which of the reactions (a, b, or c) would tend to go to completion at 25 degree C and which would reach an equilibrium condition with significant amounts of all reactants and products present?Explanation / Answer
LET ME FIRST PUT THE REACTIONS THAT ARE GIVEN
1. N2 (g) + 3H2 (g) >>>> 2NH3 (G) = -33 kj
2. 4NH3 (g)) + 5O2 (g) >>>> 4NO (g) = -1011 kj
3. N2 (g) + 2O2 (g) >>> 2NO2 (g) = 173.1 KJ
4. 2N2 (g) + O2 (g) >>>> 2N2O = +208.4 KJ
A. to get N20 + 3/2 O2 >>> 2NO2
try reversing 4 ( half it) and put eq 3 as it is
N2 + 2O2 >>>> 2NO2 = +102.6 KJ
N2O >>> N2 + 1/2 O2 = (+208.4/2) = -104.2 NOW CANCEL SIMILAR COEFFICIENTS AND SUBTRACT
= N2O + 3/2O2 >>> 2NO2 = (102.6 -(-104.2) = 206.8 kj
B, 2H2 (g) + O2 (g) >>>> 2H2O (l)
put eq 1,,half the eq 2. reverse eq 3
N2 + 3H2 >>> 2NH3 -33 KJ
2NH3 + 5/2O2 >>> 2NO + 3H20 (l) =( 1011.5/2) 505.75 kj
2NO >>>> N2 + O2 = -173.1 KJ (now cancel similar coeff, subtract )
it gives 3H2 + 3/2O2 >>>> 3H2O (l) ( -33 -505.75 -(-173.1) = -365.65 kj