The keg for the reaction
2 SO2(g) + O2(g) → 2 SO3(g) is
Keg = [SO3]^2/ {(SO2)^2 ( O2)}
Keg (equilibrium constant) is the ratio of of equilibrium concentration of the product raised to the power of their stoichiometric coefficient to the equilibrium concentration of the reactant raised to the power of their stoichiometric coefficient.
When the product formation is decreased if a substance B is added to an enzyme reaction and more substrate being added would not increase the amount of produce formed, then we assume that substance b could be a noncompetitive inhibitor. This type of inhibitor would be one that would bind to the enzyme with or without the presence of a substrate in different sites at the same time. It would change the conformation of the enzyme and also the active sites. As a result, the substrate would not be able to bind to the enzyme more effectively than the usual. The overall efficiency would decrease.
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
150/30 = 5
HF1 20/2 = 10
HF2 10/2 = 5
HF3 5/2 = 2.5
HF4 2.5/2 = 1.25
HF5 1.25/2 = 0.625
Answer: 0.63g
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