Assuming that the reaction from A and C to AC5 is only
one-step (or an elementary reaction) with a balanced chemical reaction of:
<span>A + 5 C ---> AC5 </span>
Therefore the formation constant can be easily calculated
using the following formula for formation constant:
Kf = product of products concentrations / product of reactants
concentration
<span>Kf = [AC5] / [A] [C]^5 </span>
---> Any coefficient from the balanced chemical
reaction becomes a power in the formula
Substituting the given values into the equation:
Kf = 0.100 M / (0.100 M) (0.0110 M)^5
Kf = 6,209,213,231
or in simpler terms
<span>Kf = 6.21 * 10^9 (ANSWER)</span>
Answer:
-12.5 kJ/mol
Explanation:
The free-energy predicts if a reaction is spontaneous or not. If it is, ΔG < 0. When a reaction happens by steps, the free-energy of the global reaction can be calculated by the sum of the free-energy of the steps (Hess law). If it's needed to operations at the reaction the same operation must be done in the value of ΔG (if the reaction is inverted, the signal of ΔG must be inverted).
Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol
ATP → ADP + Pi ∆G'° = –30.5 kJ/mol (x-1)
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Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol
Pi + ADP → ATP ∆G'° = 30.5 kJ/mol
The bold compounds are in opposite sides, so they'll be canceled in the sum of the reactions:
Phosphocreatine + ADP → creatine + ATP
∆G'° = -43.0 + 30.5
∆G'° = -12.5 kJ/mol
setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :
Q<K⇒The reaction moved to the right (products)
Setup 2 :
Q=K⇒the system at equilibrium
Setup 3 :
Q>K⇒The reaction moved to the left (reactants)
Umm. . . Could you list some options? :)
Answer:
Caesium atoms will form positively charged ions.
Explanation:
Due to them having one electron in their outer orbit, it is very likely that they will give that electron away to form an octet and become stable. Hence, since they lose an electron, they lose a negative charge, and in comparison to their former non-ion self, they have gained a positive charge.
Hence, they will form positively charged ions.
Hope this helped!
