Question

Calculate ∆S⁰R for the reaction 3H_{2}(g) +
N_{2}(g) = 2NH_{3}(g) at **700 K**
given the following:

S⁰_{298K} for H2g = 197.7 J K^{-1} mol
^{-1}

S⁰_{298K} (N2g) = 191.6 J K^{-1} mol
^{-1}

S⁰_{298K} (NH3g) = 130.7 J K^{-1} mol
^{-1}

Cp (H_{2}g) in J K^{-1} mol ^{-1} = 22.7
+ 4.4 x 10^{-2} T – 1.1x10^{-4} T^{2}

Cp (N_{2}g) in J K^{-1} mol ^{-1} = 30.8
– 1.2 x 10^{-2} T + 2.4 x10^{-5} T^{2}

Cp (NH_{3}g) in J K^{-1} mol ^{-1} =
29.3 + 1.1 x 10^{-2} T + 4.2x10^{-5}
T^{2}

Assume that T is in K for heat capacity. Previous solution attempt is incorrect

Answer #1

To perform the calculation of the standard entropy change at 700 K, the integral of the respective Cp / T for each substance must be added to the value of S (298K):

S 700K for H2g = 197.7 J / mol * K + ∫ 22.7 / T + 4.4 x 10-2 - 1.1x10-4 T (from 298 K to 700 K) = 197.7 + 15.01 = 212.71 J / mol * K

S 700K for N2g = 191.6 + ∫ 30.8 / T - 1.2 x 10-2 + 2.4 x10-5 T (from 298 K to 700 K) = 191.6 + 26.29 = 217.89 J / mol * K

S 700 K for NH3g = 130.7 + ∫ 29.3 / T + 1.1 x 10-2 + 4.2x10-5 T (from 298 K to 700 K) = 130.7 + 37.87 = 168.57 J / mol * K

Calculation of ΔSr:

ΔSr (700K) = n * S products - n * S reactive = 2 * 168.57 - 3 * 212.71 - 217.89 = -518.88 J / K

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

kc= 0.00592 for the reaction below at 351 K. N2 (g) + 3H2 (g)
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(g) +H2 (g)<--->2/3 NH3 (g)

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 _

For the reaction
N2(g)+3H2(g)⇌2NH3(g)
Kp = 3.80×10−3 at 253 ∘C . What is
K for the reaction at this temperature?
Enter your answer numerically.

Consider the following reaction at 25 oC:
N2(g) + 3H2(g) --> 2NH3(g)
∆Horxn= -92.6 kJ/mol and
∆Sorxn= -199
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Calculate the change in entropy of the universe,
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A mixture of 1.0 mol each of N2, H2, and
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equilibrium at 500.0 C. Kp at this temperature is 1.45 x
10-5 atm-2.
A.) What is the value of Kc at this temperature? irst
find the relationship between Kc and Kp in
terms of RT. Then substitute T = 773K and R =...

A) Consider the following reaction where Kc
= 0.159 at 723 K.
N2(g) +
3H2(g)
2NH3(g)
A reaction mixture was found to contain
4.94×10-2 moles of
N2(g),
4.07×10-2 moles of
H2(g) and
6.19×10-4 moles of
NH3(g), in a 1.00 liter
container.
The reaction quotient, Qc, equals
__________.
The reaction__________________
A. must run in the forward direction to reach
equilibrium.
B. must run in the reverse direction to reach
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C. is at equilibrium.
B) Consider the following reaction where Kc...

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N2 (32 bar, 1500 K). Assume N2 to be an ideal
gas. CP = 28.58 + 3.77 x 10-3 T – 0.50 x
105 T-2 J K-1 mol-1

1.) The equilibrium constant for the chemical equation
N2(g)+3H2(g) <-->
2NH3(g)
is Kp = 1.09 at 209 °C. Calculate the value of the Kc for the
reaction at 209 °C.
2.) At a certain temperature, 0.3411 mol of N2 and 1.581 mol of
H2 are placed in a 1.50-L container.
N2(g)+3H2(g) <-->
2NH3(g)
At equilibrium, 0.1801 mol of N2 is present. Calculate the
equilibrium constant, Kc.
3.) At a certain temperature, the Kp for the decomposition of
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Consider the reaction: N2(g) + 3H2(g) → 2NH3(g). At a
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(b) A what rate (in M/s) is NH3 being produced? {3 points}
(c) Are these average or instantaneous rates? {1 point}
[3] A certain first-order reaction has a rate constant of 1.65
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