Keep in mind that when two moles of potassium chlorate are decomposed, 3 moles of oxygen are produced.Think of the coefficients: since the mole ratio is 2 to 3, it means that<span> 2 mol KClO3 will turn into 3 mol O2.
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2kclO3->2kcl+3O2
Answer:
- What is the AGⓇ of this reaction? 0.
- Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
- What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
- If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.
Explanation:
1. To calculate the delta G of a reaction given the K, we use the following equation:
ΔG°= -RT ln K.
Which gives us 0 when K is 1.
2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.
3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.
4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.
<h2>Question:- </h2>
A solution has a pH of 5.4, the determination of [H+].
<h2>Given :- </h2>
- pH:- 5.4
- pH = - log[H+]
<h2>To find :- concentration of H+</h2>
<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>
<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>
Take negative to other side
-pH = log H+
multiple Antilog on both side
(Antilog and log cancel each other )
Antilog (-pH) = [ H+ ]
New Formula :- Antilog (-pH) = [+H]
Now put the values of pH in new formula
Antilog (-5.4) = [+H]
we can write -5.4 as (-6+0.6) just to solve Antilog
Antilog ( -6+0.6 ) = [+H]
Antilog (-6) × Antilog (0.6) = [+H]
put the value in equation
![{10}^{ - 6} \times 4 = [H+] \\ 4 \times {10}^{ - 6} = [H+]](https://tex.z-dn.net/?f=%20%7B10%7D%5E%7B%20-%206%7D%20%20%20%5Ctimes%204%20%3D%20%5BH%2B%5D%20%5C%5C%204%20%5Ctimes%20%20%20%7B10%7D%5E%7B%20-%206%7D%20%20%3D%20%5BH%2B%5D)
