Using Differenatial rate law expression
1. For the reaction 4NH3(g) + 302(g) → 2N2(g) + 6H2O(g) it was found that at...
Given the system: 4NH3(g) + 302(g) = 2N2(g) + 6H2O(1) AH=-1530.4 kJ a) Discuss (and provide reasoning) how the ammonia (NH3) concentration will be affected at equilibrium: i) Removing O2: ii) Adding additional water: iii) Increasing volume:
6 (13 pts) Given the system: 4NH3(g) + 302(g) = 2N2(g) + 6H2O(1) AH=-1530.4 kJ a) Discuss and provide reasoning) how the ammonia (NH3) concentration will be affected at equilibrium: i) Removing 02: ii) Adding additional water: iii) Increasing volume:
. Given the following reaction at 25o C. 4NH3(g) + 5 O2(g) 4NO(g) + 6H2O(l) ΔHo = -1168 kJ ΔHo f(NH3) = -46.2 kJ/mol; ΔHo f(H2O) = -285.8 kJ/mol. What is the standard enthalpy of formation of NO gas at 25o C?
Consider the following reaction: 4NH3 + 5O2 4NO + 6H2O Determine the following rates at a time when the rate of consumption of O2 is 1.26e-03 M/s. a) rate of consumption of NH3 M/s b) rate of formation of NO M/s c) rate of formation of H2O M/s d) rate of reaction M/s
Determine delta sub r H in Kj/mol for this reaction. 4No+6H2O --> 4NH3 +5O2 using the equations and the enthalpy change of the reactions given. N2 + O2 --> 2NO delta sub r H= 180.1 NH3 --> 1/2 N2 +1.5H2. delta sub r H= 54.3 2H2O --> 2H2 + O2 delta sub r H= 486.7
2) 4NH3(g) + 302(g) + 6H2O(l) + 2N2(g) (a) 2 moles of NH, and 3 moles of Oz will produce ? moles Nz? (b) 5.0 g of NH, and 5.0 g of Oz will produce ? g of H2O? How many H2O molecules? 3) A mixture of 80.0 g of chromium (VI) oxide, Cr2Os, and 8.00 g of solid carbon, C, is used to produce elemental chromium, Cr, by the reaction below Cr2O3(aq) + 3C (s) → 2Cr (s) +...
The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0160 mol gaseous NH3 and 0.0220 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.70×10-3 M. Calculate Keq for the reaction at this temperature.
Confused on #5
5) Calculate the enthalpy of the following reaction: N2 + O2 ---> 2NO Given: 4NH3 + 502 ---> 4NO + 6H2O AH° = -1170 kJ 2N2 + 6H2O ---> 4NH3 + 302 AH° = +1530 kJ Solution:
Calculate ΔGo for the reaction of ammonia with oxygen: 4NH3 (g) + 3 O2 (g) → 2N2 (g) + 6 H2O (g) using the following data: Substance NH3 (g) O2 (g) N2 (g) H2O (g) ΔGo (kJ/mol) - 16 0 0 -228.6 a) 212.6 kJ b) 1307.6 kJ c) - 212.6 kJ d) -1435.6 kJ e) -1307.6kJ
Consider the two reactions. 2NH3(g)+3N2O(g)4NH3(g)+3O2(g)⟶4N2(g)+3H2O(l)⟶2N2(g)+6H2O(l) Δ?∘=−1010 kJΔ?∘=1531 kJ2NH3(g)+3N2O(g)⟶4N2(g)+3H2O(l) ΔH∘=−1010 kJ4NH3(g)+3O2(g)⟶2N2(g)+6H2O(l) ΔH∘=1531 kJ Using these two reactions, calculate and enter the enthalpy change for the reaction below. N2(g)+12O2(g)⟶N2O(g)N2(g)+12O2(g)⟶N2O(g)