N2 (g) + 3 H2 (g) - 2 NH3 (g), Ke = 112 (at 450 K) AH° = + 29 kJmol In this reaction at equilibrium, the concentration of N2 is found to 0.040 M, and that of NH3 is 0.251 M. what is the concentraton of H2 ? 0.211 M b. 0.242 M c. 0.120 M d. 0.065 M e. 0.291 M
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
35) please see attached for the endothermic reaction
![**Chemical Equilibrium Problem**
The given chemical equilibrium reaction is as follows:
\[ \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \]
Where the equilibrium constant \( K_c \) is 112 at a temperature of 450 K and the change in enthalpy (\( \Delta H^\circ \)) is +29 kJ/mol.
In this reaction at equilibrium, the concentration of \(\text{N}_2\) is found to be 0.040 M, and the concentration of \(\text{NH}_3\) is 0.251 M. The question asks for the concentration of \(\text{H}_2\).
**Options provided:**
a. 0.211 M
b. 0.242 M
c. 0.120 M
d. 0.065 M
e. 0.291 M
**Explanation:**
To solve for the concentration of \( \text{H}_2\):
1. Write the expression for the equilibrium constant (\( K_c \)):
\[ K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]
2. Substitute the known values into the equation:
\[ 112 = \frac{(0.251)^2}{(0.040)[\text{H}_2]^3} \]
3. Simplify and solve for \([\text{H}_2]\):
\[ 112 = \frac{0.063001}{0.040 [\text{H}_2]^3} \]
\[ 112 \times 0.040 [\text{H}_2]^3 = 0.063001 \]
\[ 4.48 [\text{H}_2]^3 = 0.063001 \]
\[ [\text{H}_2]^3 = \frac{0.063001}{4.48} = 0.01406317 \]
\[ [\text{H}_2] = (0.01406317)^{\frac{1}{3}} \approx 0.240 M \]
Thus, the concentration of \(\text{H}_2\) is approximately 0.242 M (option b).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48deb947-3d41-43de-ac43-bef03e01cc59%2F185d3aa6-6a5d-4102-ab2a-d623ac87b10c%2Fbw1k5mr.png&w=3840&q=75)
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