Answer:
1) No shift
2) No shift
3) Leftward shift
4)Rightward sifht
Explanation:
1) 2) Adding N or Removing N in the equilibrium will produce No shift, because of its solid state, the N is not contemplated in the equilibrium equation:
![K=\frac{[L]^2}{[M]^3}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BL%5D%5E2%7D%7B%5BM%5D%5E3%7D)
3) Increasing the volume produces a decrase in the preassure due to the expansion of the gases. This will cause a leftward shift, because the system will try to increase the moles of gas and in consecuence of this, also increase the preassure.
4) Decreasing the volume has the opposite effect of the item 3): the preassure will increase and the system will consume moles of gas to decrease it, producing a rightward shift.
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
It is c I hope I helped out with this question!.
To know the answer, you either know what is really the
nature and chemistry of a sugar solution. You can also know the answer by
knowing the meaning of entropy. Entropy is often interpreted as the degree of
disorder or randomness in the system. So the correct statement is that the
system becomes more disordered and has an increase in entropy.
