4...........................
Answer:
At 430.34 K the reaction will be at equilibrium, at T > 430.34 the
reaction will be spontaneous, and at T < 430.4K the reaction will not
occur spontaneously.
Explanation:
1) Variables:
G = Gibbs energy
H = enthalpy
S = entropy
2) Formula (definition)
G = H + TS
=> ΔG = ΔH - TΔS
3) conditions
ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction
4) Assuming the data given correspond to ΔH and ΔS
ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K
=> T = [ΔH - ΔG] / ΔS
ΔG = 0 => T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K
This is, at 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.
Answer: 1:4.69
Explanation:
The ratio can be expressed as:
Ua/Ub= √(Mb/Ma)
Where Ua/Ub is the ratio of velocity of hydrogen to carbon dioxide and Ma is the molecular mass of hydrogen gas= 2
Mb is the molecular mass of CO2 = 44
Therefore
Ua/Ub= √(44/2)
Ua/Ub = 4.69
Therefore the ratio of velocity of hydrogen gas to carbon dioxide = 1:4.69
which implies hydogen is about 4.69 times faster than carbon dioxide.
<span>Data:
pH = 5.2
[H+] = ?
Knowing that: (</span><span>Equation to find the pH of a solution)</span>
![pH = -log[H+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%2B%5D)
<span>
Solving:
</span>
![5.2 = - log [H+]](https://tex.z-dn.net/?f=5.2%20%3D%20-%20log%20%5BH%2B%5D)
Knowing that the exponential is the opposite operation of the logarithm, then we have:
![[H+] = 10^{-5.2}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%2010%5E%7B-5.2%7D)
![\boxed{\boxed{[H+] = 6.30*10^{-6}}}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5BH%2B%5D%20%3D%206.30%2A10%5E%7B-6%7D%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
