Question requires a change resulting in an increase in both forward and reverse reactions. Now lets discuss options one by one and see there impact on rate of reactions.
1) <span>A decrease in the concentration of the reactants:
When concentration of reactant is decreased it will shift the equilibrium in Backward direction, so resulting in increasing the backward reaction and decreasing the forward direction. Hence, this option is incorrect.
2) </span><span>A decrease in the surface area of the products:
Greater the surface Area greater is the chances of collision and greater will be the rate of reaction. As the surface area of products is decreased it will not favor the backward reaction. Hence again this statement is incorrect according to given statement.
3) </span><span>An increase in the temperature of the system:
An increase in temperature will shift the reaction in endothermic side. Hence, if the reaction is endothermic, an increase in temperature will increase the rate of forward direction or if the reaction is exothermic it will increase the rate of reverse direction. Hence, this option is correct according to given statement.
4) </span><span>An increase in the activation energy of the forward reaction:
An increase in Activation energy will decrease the rate of reaction, either it is forward or reverse. So this is incorrect.
Result:
Hence, the correct answer is,"</span>An increase in the temperature of the system".
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
Answer:
Ca, Zn, Ni, Al, Fe
Explanation:
The activity series is an arrangement of metals in order of decreasing reactivity.
Metals that are high up in the activity series are very reactive. They displace metals that are lower in the activity series from an aqueous solution of their salts.
The metals; Ca, Zn, Ni, Al, Fe are all above Pb in the activity series hence they will cause Pb2+ ions to come out of solution as solid Pb.
The answer will be 70 pascal as pressure =.F upon area
