Answer:
![K=\frac{[CaO][CH_{4}][H_{2}O ]^{2} }{[CaCO_{2}][H_{2}]^4 }](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCaO%5D%5BCH_%7B4%7D%5D%5BH_%7B2%7DO%20%5D%5E%7B2%7D%20%20%7D%7B%5BCaCO_%7B2%7D%5D%5BH_%7B2%7D%5D%5E4%20%20%7D)
Explanation:
The equilibrium expression is the K value equal to the product of the concentrations of the products over the product of the concentrations of the reactants. If there is a coefficient in front of the compound, raise the molecule to that power.
Since K is big, more product is expected. This is because of mathematic principles. A large numerator with a small denominator will produce a large number.
Mole is mass (g) / Molar mass (mole/gram)
So to find mass in gram multiply the no.mole by Molar mass
Melted rock and minerals below the earth's crust is also known as magma
Answer:
Explanation:
A bronsted lowry acid just means that it donates a proton.
An arrhenius acid dissolves in water to donate a proton
the only difference is that an arrhenius acid must dissolve in water but it still donates a proton so it is considered a bronsted lowry acid
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
