We are given with
4.35 g Phosphoric acid
5.25 g KOH
3.15 g K3PO4 produced
The reaction is
H3PO4 + 3KOH => K3PO4 + 3H2O
First, convert masses into moles.
Then, determine the limiting reactant.
Next, determine the maximum amount of K3PO4 that can be produced from the limiting reactant.
Lastly, calculate the percent yield by dividing the actual amount produced by the theoretical amount produced.
Answer:
The value of the of ΔG for the new reaction will be same as the given value of -20kcal/mol.
Explanation:
In an enzyme-catalyzed reaction, the concentration or amount of enzyme will not affect the equilibrium constant of the reaction due to which ΔG for the reaction will remain unaffected. Here enzymes are acting as a catalyst that cannot alter law thermodynamics and equilibrium of the reaction.
Since the enzyme amount will not affect the equilibrium of the reaction, the value of ΔG will remain the same as given -20 kcal/mol.
Answer
- continuous removal of PH3
- adding more of P into the system
Explanation:
In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);
- The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:
heat+A⇌BΔH=+
- Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.
- According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.
The formula to be used for this problem is as follows:
E = hc/λ, where h is the Planck's constant, c is the speed of light and λ is the wavelength. Also 1 aJ = 10⁻¹⁸ J
0.696×10⁻¹⁸ = (6.62607004×10⁻³⁴ m²·kg/s)(3×10⁸ m/s)/λ
Solving for λ,
λ = 2.656×10⁻⁷ m or <em>0.022656 nm</em>
C. The salt bridge maintains the flows of ions and allows electrons to move from the anode to the cathode.
