Answer:
Option A
Explanation:
A) Yes. The reaction reaches equilibrium when the rate of reaction of the reverse reaction is equal to the rate of the forward reaction , then the only cause for the reverse reaction to be favoured is that the initial rate of the reverse was greater than the forward one.
B) No. The rate constant of the reverse reaction can be greater than the forward one but the rate also depends on concentrations, thus a reverse reaction with greater rate constant can result in the net reaction proceeding in the forward reaction, the reverse reaction or be at equilibrium depending on the concentrations or reactants and products
C) No. A lower activation energy means a higher rate constant , but a higher rate constant does not mean that the net reaction will proceed to the reactants ( see point B)
D) No. The energy changes determine conditions under thermodynamic equilibrium and therefore the net direction of the reaction will depend on the temperature and concentrations of reactants and products with respect to the equilibrium conditions.
Answer:
C. 3CO(g) + Fe2O3(s)
Explanation:
The substance(s) to the hath left of the arrow in a chemical equation art hath called reactants. A reactant is a substance yond is presenteth at the starteth of a chemical reaction. The substance(s) to the right of the arrow art hath called products. A product is a substance yond is presenteth at the endeth of a chemical reaction
So in this example, 3CO(g) + Fe2O3(s) art the reactants.
The 2Fe(S) + 3CO2(G) art the products.
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Answer:
6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.
Explanation:
![Rate = k[AB]^2](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BAB%5D%5E2)
The order of the reaction is 2.
Integrated rate law for second order kinetic is:
![\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_t%5D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%2Bkt)
Where, ![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration = 1.50 mol/L
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the final concentration = 1/3 of initial concentration = 
= 0.5 mol/L
Rate constant, k = 0.2 L/mol*s
Applying in the above equation as:-
<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>
Thermoplastics and thermosetting polymers Examples include: polyethylene (PS) and polyvinyl choline (PVC). Common thermoplastics range from 20,000 to 50,000 amu, while thermosets are assumed to have infinite molecular weight.
