For the reaction N2(g) + 3 H2(g)→2 NH3(g), ∆rH=−45.94kJ/mol
At 298K,∆rG=−32.8kJ/mol
Estimate∆rG of the same reaction at 0C.
ΔG = -32.8 KJ/mol
ΔH = -45.94 KJ/mol
T = 298 K
use:
ΔG = ΔH - T*ΔS
-32.8 = -45.94 - 298.0 *ΔS
ΔS = -0.0441 KJ/mol.K
ΔS = -44.094 J/mol.K
Now we have:
ΔH = -45.94 KJ/mol
ΔS = -44.094 J/mol.K
= -0.04409 KJ/mol.K
T= 0.0 oC
= (0.0+273) K
= 273 K
use:
ΔG = ΔH - T*ΔS
ΔG = -45.94 - 273.0 * -0.0441
ΔG = -33.9023 KJ/mol
Answer: -33.9 KJ/mol
For the reaction N2(g) + 3 H2(g)→2 NH3(g), ∆rH=−45.94kJ/mol At 298K,∆rG=−32.8kJ/mol Estimate∆rG of the same reaction...
For the reactionN2(g) + 3 H2(g)−−→2 NH3(g),∆rH=−45.94kJ/mol.At 298K,∆rG=−32.8kJ/mol. Estimate ∆rG of the same reaction at 0C. Use Gibbs-Helmholtz equation.
Find ΔrG for the following (in kJ mol-1) N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) The conditions for this reaction are: Temp: 298k P - NH3 = 0.95 bar P - H2 = 1.95 bar P - N2 = 1.25 bar NH3(g) ?H ∙(kJ mol-1) = -45.9 ?G ∙(kJ mol-1) = -16.4 S ∙(J K-1 mol-1)192.8 N2(g) ?H ∙(kJ mol-1) = 0 ?G ∙(kJ mol-1) = 0 S ∙(J K-1 mol-1)191.6 H2(g) ?H ∙(kJ mol-1) = 0...
3. (20 points) For the reaction N2(g) + 3H2(g) - kJ/mol. Estimate A, G of the same reaction at 0 °C. 2 NH3(g) at 298K, A,Gº = -32.8
N2 (g) + 3 H2 (g) -> 2 NH3 (g) assume 0.210 mol N2 and 0.674 mol H2 arw present initially. after complete reaction, how many moles of ammonia are produced? how many moles of H2 remain? how many moles of N2 remain?
Consider the reaction: N2(g) + 3 H2(g) « 2 NH3(g) a. Write the expression for the equilibrium constant, K, for this reaction. b. An equilibrium misture of N2, H2, and NH3 at 300°C is analyzed, and it is found that: [N2] = 0.25 mol/L, [H2] = 0.15 mo/L, and [NH3] = 0.090 mol/L. Find K at 300°C for this reaction.
Using the equations N2 (g) + 3 H2 (g) → 2 NH3 (g) AH° = -91.8 kJ/mol C(s) + 2 H2 (g) → CH4 (g) AH° = -74.9 kJ/ mol H2 (g) + 2 C(s) + N2 (g) → 2 HCN (g) AH° = 270.3 kJ/mol Determine the enthalpy for the reaction CH4 (g) + NH3 (g) → HCN (g) + 3 H2 (g).
At 400 K, the reaction N2 (g) + 3 H2 (g) → 2 NH3 (g) reaches equilibrium when the partial pressures of nitrogen, hydrogen, and ammonia gases are 4.00 atm, 1.00 atm, and 1.05 x 10−2 atm, respectively. Given that the standard enthalpy of the reaction at 400K is DH = -94 kJ/mol, estimate the value of the equilibrium constant KP at 450 K assuming that the standard enthalpy of reaction does not vary significantly with temperature in this temperature...
Consider the reaction: 2 NH3 (g) → 3 H2 (g) + N2 (g) Which of the following is true about entropy associated with this reaction? A. The entropy of the system will be positive because the mol of gas is increasing. B. The entropy of the system will be negative because the mol of gas is decreasing. C. The entropy of the system will be negative because the mol of gas is increasing. D. The entropy of the system will...
I will rate. thank you N2(g)+ 3 H2(g)= 2 NH3 Find delta G of reaction when mixing 4 mol of nitrogen, 12 moles of of h2 and 4 mol nh3 (pressure = 1 bar) temp = 50c
1. For the reaction 3 H2(g) + N2(g) → 2 NH, (g), 3 mol H2 is reacted with 6 mol N2- mol of NH3 is produced mol of Hy remains _mol Ny remains 2. For the reaction 2 N H (1) + N20 (1) 3N2(g) + 4 H 0 (), 160 g N,H, is mixed with 160 g N204 (Answer: 125 g H20) is the limiting reagent _g H2O is produced 3. For the reaction Fe203 (s) + 3 CO(g)...