Answer: D
Explanation: I got it wrong on my test :( and got this answer as correct.
The complete balanced chemical equation for this is:
<span>3KOH + H3PO4
--> K3PO4 + 3H2O</span>
First we calculate the number of moles of H3PO4:
moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol
From stoichiometry, 3 moles of KOH is required for every
mole of H3PO4, therefore:
moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole
H3PO4) = 0.0468 mol
Calculating for volume given molarity of 0.350 M KOH:
Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7
mL
Answer:
<span>133.7 mL KOH</span>
The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.
<h3>Gibb's free energy of reactions</h3>
It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;
On this note, it follows that;
Hence, the Gibb's free energy for the reaction is;
- ΔG = 14.6 - 71.52
- ΔG = -56.92KJ
Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.
Read more on Gibb's free energy;
brainly.com/question/13765848
From the given observations,
You can see that as the concentration is doubled, half-life is halved.
That is,half-life is inversely proportional to concentration
As t( half-life) ~ 1/a^(n-1)
For this case n = 2,second order reaction.
R = k X a^n
Using the above formula you will get the rate and rate constant.
