We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
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Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
Answer:
The balanced equation for this reaction will be
→ 
We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2
So using the unitary method we will get that
- 1 mole of CH4 → 4 mole of 4 mole of fluorine
- 0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction
but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.
- 4 moles of fluorine → 1 mole of CF4
- 0.56 mole →
= 0.14mole of CF4
- 4 moles of fluorine → 4 moles of HF
- 0.56 mole of fluorine → 0.56 mole of HF
now to find the heat released we have the formula as
DELTA H = n * Delta H of product - n *delta H of reactant
where n is the moles of the reactant and product.
note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.
Answer:
a. 211.7
Explanation:
Iron Pyrite reacts with Oxygen to produce Iron (II) Oxide and Sulphur (IV) Oxide.
The equation is as follows:
4FeS₂₍s₎ + 11O₂₍g₎ → 2Fe₂O₃₍s₎ + 8SO₂₍g₎
From the equation, 4 moles of FeS₂ produce 8 moles of SO₂.
Therefore the reaction ratio is 4:8 or 1:2
198.20 grams of FeS₂ into moles is calculated as follows:
Moles= Mass/RMM
RMM of FeS₂ is 119.9750g/mol.
Number of moles = 198.20/119.9750g/mol
=1.652 moles of FeS₂
The reaction ratio of FeS₂ to SO₂ produced is 1:2
Thus SO₂ produced = 1.652 moles×2/1=3.304 moles
The mass of SO₂ produced =Moles ×RMM
=3.304 moles ×64.0638 g/mol
=211.667 grams
=211.7g
Hello.
He have that:
[H+][OH-] = 10⁻¹⁴
3.64 x 10⁻⁸ [OH-] = 10⁻¹⁴
[OH-] = 0.27 x 10⁻⁶ mol/L
Are you gonna list the ppl?
