Question

Use the following information to find Δ*H*°_{f}
of gaseous HCl:

N_{2}(*g*) + 3H_{2}(*g*) →
2NH_{3}(*g*)
Δ*H*°_{rxn} = - 91.8 kJ

N_{2}(*g*) + 4H_{2}(*g*) +
Cl_{2}(*g*) →
2NH_{4}Cl(*s*)
Δ*H*°_{rxn} = - 628.8 kJ

NH_{3}(*g*) + HCl(*g*) →
NH_{4}Cl(*s*)
Δ*H*°_{rxn} = - 176.2 kJ

Answer #1

N2(g) + 3H2(g) → 2NH3(g) ΔH°rxn = - 91.8 kJ 1equation

N2(g) + 4H2(g) + Cl2(g) → 2NH4Cl(s) ΔH°rxn = - 628.8 kJ 2 equation

NH3(g) + HCl(g) → NH4Cl(s) ΔH°rxn = - 176.2 kJ 3 equation

Multiply the third equation by -2. The (-) will reverse the
direction of the reaction.

Multiply the first reaction by -1.

Sum the modified first and third and add the second. You will be
left with

H2(g) + Cl2(g) ---> 2HCl(g).

Now perform the same operations on the ΔH. e.g., third value times
-2, first value times -1. Add them now and add the second value.
That's it.

(-2)*(-176.2) + (-1)*(-91.8) + 1*(-628.8) = -184.6/2 =so, ∆HRx =
−92.3 kJ mol−1 of HCl

N2 (g) + 3H2 (g) → 2NH3(g) ΔH =
-92.2 kJ
What mass of ammonia is theoretically produced if the above
reaction released 183 kJ of heat?
Please show work

Applying Hess’s Law, from the enthalpies of reactions,
N2(g) + 3H2(g) → 2 NH3(g) ΔH = − 91.8 kJ
O2(g) + 2H2(g) → 2H2O (g) ΔH = − 483.7 kJ
N2(g) + O2(g) → 2NO(g) ΔH = 180.6 kJ
Calculate the enthalpy change (ΔHrxn) for the reaction: 4NH3(g)
+ 5O2(g) → 4 NO (g) + 6H2O(g)

Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f
for IF: ΔH°f (kJ/mol) IF7(g) + I2(g) → IF5(g) + 2 IF(g) ΔH°rxn =
-89 kJ IF7(g) -941 IF5(g) -840

Determine the equilibrium constant Kp at 25°C for the reaction:
N2(g) + 3H2(g) ⇆ 2NH3(g)
(G°f (NH3(g)) = -16.6 kJ/mol)

Use the ΔH°f and ΔH°rxn information provided to calculate ΔH°f
for IF: ΔH°f (kJ/mol) IF7(g) + I2(g) → IF5(g) + 2 IF(g)
ΔH°rxn = -89 kJ
IF7(g) -941
IF5(g) -840
Answers:
(a) -190 KJ/mol
(b) 101 KJ/mol
(c) 24 KJ/mol
(d) -95 KJ/mol
(e) -146 KJ/mol

NO2(g) + 7⁄2H2(g) ---> 2H2O(ℓ) + NH3(g)
ΔH° = ??? kJ
Using the following two equations:
2NH3(g) ---> N2(g) + 3H2(g)
ΔH° = +92 kJ
1⁄2N2(g) + 2H2O(ℓ) ---> NO2(g) + 2H2(g)
ΔH° = +170 kJ

Please solve step by step for Part D
Consider the Haber synthesis of gaseous NH3
(ΔH∘f=−46.1kJ/mol ;ΔG∘f=−16.5kJ/mol) :
N2(g)+3H2(g)→2NH3(g)
Part A
Use only these data to calculate ΔH∘ and ΔS∘
for the reaction at 25 ∘C.
Express your answers using three significant figures separated
by a comma.
ΔH∘, ΔS∘ =
-92.2,-199
kJ, J/K
Part D. What are the equilibrium constants Kp
and Kc for the reaction at 370 K ? Assume that
ΔH∘ and ΔS∘ are independent of
temperature.
Express...

calculate the standard free energy change delta G for reaction
N2 (g) +3H2(g)—>2NH3
N2 delta H=0.00kj mol^-1s=+191.5J mol^-1K^-1
H2 delta H=0.00kj mol^-1,s = +130.6j mol^-1 k-1
NH3 delta H=-46.0kj mol^-1,s =192.5 J mol^-1 k-1
A. +112.3 kJ
B.-87.6kJ
C.-7.4kJ
D.-32.9 kJ
E.-151.1kJ

Ammonia is formed by the Haber process according to the
following reaction:
N2(g) + 3H2(g) ⇌ 2NH3(g)
Use the following data table to answer the questions below:
Substance: ΔHf (kJ/mol) So (J/(mol*K)
N2(g) 0 187.4
H2(g) 0 127.1
NH3(g) -47.3 197.6
Part 1:
Using the table in the introduction, calculate the value of ΔH
in units of kJ/mol. After, calculate the value of ΔS in units of
J/(mol*K). Finally, cCalculate the value of ΔG in units of kJ/mol
for...

consider the following reaction:
2NH3 (g) <=> N2 (g) + 3H2
(g)
if 7.92 x 10-4 moles of NH3, 0.336 moles
of N2 and 0.287 moles of H2 are at
equilibrium in a 10.2 L container at 884 K, the value of the
equilibrium, Kp, is _

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 20 minutes ago

asked 33 minutes ago

asked 33 minutes ago

asked 37 minutes ago

asked 43 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 3 hours ago

asked 3 hours ago