. 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?
We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
. Given the following reaction at 25o C. 4NH3(g) + 5 O2(g) 4NO(g) + 6H2O(l)...
Use standard enthalpies of formation to determine the ΔHo in kJ for the following reaction: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) ΔHfo (NH3(g)) = -45.90 kJ/mol ΔHfo (NO(g)) = 90.29 kJ/mol ΔHfo (H2O(g)) = -241.83 kJ/mol
Consider the following equilibrium reaction: 4NH3(g) + 5O2(g) <---> 4NO(g) + 6H2O(g) Heat = -904.4 kJ How does each of the following changes affect the yield of NO at equilibrium? Answer increase, decrease, or no change. a. increase [NH3] b. Increase [H2O] c. Decrease [O2] d. Decrease the volume of the container in which the raction occurs. e. Add a catalyst f. Increase temperature
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
A scientist measures the standard enthalpy change for the following reaction to be -2847.0 kJ: 2C2H6(g) + 7 02(g) —4CO2(g) + 6H2O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CzH6(g) is kJ/mol. A scientist measures the standard enthalpy change for the following reaction to be -15.0 kJ : Ca(OH)2(aq) + 2 HCl(aq) +CaCl(s) + 2 H20(1) Based on this value and the standard enthalpies of formation for...
Consider the following reaction: 4NH3(g) + 502(g) – 4NO(g) + 6H2O(g); AH = -906 kJ Now answer the following two questions. 1. Which condition will release more heat? A. Combining 4.0 g of NH3 with excess O2 or B. Combining 5.0 g of O2 with excess NH3 2. How much heat will be released from your answer in question 1? Condition will release more heat and the value of this heat energy is
Consider the following reaction: 4NH3 + 502 --> 4NO + 6H2O. In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of 02. Only 0.490 g of NO is actually produced. What is the percent yield of NO?
Consider the reaction 4NH3(g) +502(9) +4NO(g) + 6H2O(9) for which AH° = -905.2 kJ and A Sº = 180.5 JK at 298.15 K. (1) Calculate the entropy change of the UNIVERSE when 1.669 moles of NH3(g) react under standard conditions at 298.15 K. = A Suniverse JK (2) Is this reaction reactant or product favored under standard conditions? (3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the...
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:
(NO(g)) = 20. Determine the enthalpy for the following reaction, given AH (NH3(g)) = -46.1 kJ/mol, AH +90.3 kJ/mol, and AH (H2O(g)) = -241.8 kJ/mol. 4NH3(g) + 502(g) → 4NO(g) + 6H2O(g) AH x = ? kJ a. -1274 kJ/mol d. -905.2 kJ/mol b. -1,996 kJ/mol e. -105.4 kJ/mol c. +1,274 kJ/mol