Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
<h3>Rate = k×[A]</h3>
b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
<h3>k = 0.213s⁻¹</h3>
When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant.
Let's do the ICE analysis
2 NH₃ ⇄ N₂ + 3 H₂
I 0 1.3 1.65
C +2x -x -3x
-------------------------------------
E 0.1 ? ?
The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:
Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 molEquilibrium N₂ = 1..3 - 0.05 = 1.25 mol
For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is
![K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}](https://tex.z-dn.net/?f=%20K_%7BC%7D%20%3D%20%5Cfrac%7B%5B%20C%5E%7Bc%7D%20%5D%7D%7B%5B%20A%5E%7Ba%7D%20%5D%5B%20B%5E%7Bb%7D%20%5D%7D%20)
Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.
Answer:
1.1 × 10⁻⁴ M
Explanation:
Let's consider the following double displacement reaction.
CuCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s)+ Cu(NO₃)₂(aq)
We can establish the following relations:
- The molar mass of AgCl is 143.32 g/mol.
- The molar ratio of AgCl to CuCl₂ is 2:1
The moles of CuCl₂ that reacted to produce 7.7 mg of AgCl are:

The molarity of CuCl₂ is:

The balanced chemical reaction is:
Zn + 2AgNO3 = Zn(NO3)2 + 2Ag
To determine the amount of the reactant left, we have to determine which is the limiting and the excess reactant. We do as follows:
5.65 g Zn ( 1 mol / 65.38 g) = 0.09 mol Zn
21.6 g AgNO3 (1 mol / 169.87 g) = 0.13 mol AgNO3
The limiting reactant would be silver nitrate since it is consumed completely in the reaction. The excess reactant would be zinc.
Excess Zinc = 0.09 mol Zn - (0.13 / 2) mol Zn = 0.025 mol Zn left