Answer:
1: At temperatures below 542.55 K
2: At temperatures above 660 K
Explanation:
Hello there!
In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:
Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:
1:
It means that at temperatures lower than 542.55 K the reaction will be spontaneous.
2:
It means that at temperatures higher than 660 K the reaction will be spontaneous.
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Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much
<span>actual delta H = 0.34 moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>
<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>
I think 5.50 M x 35.0 mL x molar mass of RbOH = mass (g)
