Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
Answer:
Oxidized and reducing agent: manganese.
Reduced and oxidizing agent: mercury.
Explanation:
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In this case, for the reaction:

We keep in mind that the species that increase the oxidation state is the oxidized one whereas the one that decrease the oxidation state is the reduced one; therefore manganese is the oxidized one as well as the reducing agent because it goes from 0 to +2 and mercury the reduced one as well as the oxidizing agent because it goes from 2+ to 0.
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Answer:
A. The reaction will proceed forward forming more CH4
B. The reaction will proceed forward forming more CH4
C. Since the reaction is exothermic, raising the temperature will cause the reaction to proceed backward, thus forming C and H2.
D. Lowering the volume makes the gas particles to be more close together thereby enhancing their collisions leading to reaction. Therefore the reaction will proceed forward forming more CH4
E. Catalyst only reduce the activation energy so the reaction can proceed faster. The reaction will proceed forward forming.
F. The following will favour CH4 at equilibrium
i. Catalyst to the reaction mixture,
ii. Both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture
iii. Adding more C to the reaction mixture.