Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation:
El arroz si produce electricidad:)
"The reaction will absorb energy" is the best conclusion according to the energy diagram of the chemical reaction.
<u>Option: B</u>
<u>Explanation:</u>
The chemical bonds in the reactions are broken and formed as per process and contributed by three major steps: reactants, transition phase and product formation. Here transition phase is in equilibrium stage drived by activation energy, where bond is partially formed and partially broken, located at higher energy level then the starters.
The reactant's energy level is less relative to the products as seen in the endothermic reactions' energy diagram, which depicts that the products are less balanced than reactants. Here when the reaction is forced to the forward direction, then it direct towards the more unbalance entities. As energy is absorbed in the endothermic reaction from surrounding, thus the enthalpy change (ΔH) for the reaction is positive.
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M