Answer:
rate = k[A][B] where k = k₂K
Explanation:
Your mechanism is a slow step with a prior equilibrium:
![\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrcl%7D%5Ctext%7BStep%201%7D%3A%26%20%5Ctext%7BA%20%2B%20B%7D%20%26%20%5Cxrightarrow%20%5Bk_%7B-1%7D%5D%7Bk_%7B1%7D%7D%20%26%20%5Ctext%7BC%7D%5C%5C%5Ctext%7BStep%202%7D%3A%20%26%20%5Ctext%7BC%20%2B%20A%7D%20%26%20%5Cxrightarrow%20%5B%20%5D%7Bk_%7B2%7D%7D%20%26%20%5Ctext%7BD%7D%5C%5C%5Ctext%7BOverall%7D%3A%20%26%20%5Ctext%7B2A%20%2B%20B%7D%20%26%20%5Clongrightarrow%20%5C%2C%20%26%20%5Ctext%7BD%7D%5C%5C%5Cend%7Barray%7D)
(The arrow in Step 1 should be equilibrium arrows).
1. Write the rate equations:
![-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]](https://tex.z-dn.net/?f=-%5Cdfrac%7B%5Ctext%7Bd%5BA%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-%5Cdfrac%7B%5Ctext%7Bd%5BB%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-k_%7B1%7D%5B%5Ctext%7BA%7D%5D%5B%5Ctext%7BB%7D%5D%20%2B%20k_%7B1%7D%5B%5Ctext%7BC%7D%5D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BC%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B1%7D%5B%5Ctext%7BA%7D%5D%5B%5Ctext%7BB%7D%5D%20-%20k_%7B2%7D%5B%5Ctext%7BC%7D%5D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BD%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B2%7D%5B%5Ctext%7BC%7D%5D)
2. Derive the rate law
Assume k₋₁ ≫ k₂.
Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.
In an equilibrium, the forward and reverse rates are equal:
k₁[A][B] = k₋₁[C]
[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)
rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]
The rate law is
rate = k[A][B] where k = k₂K
Answer:
5.645 × 10⁻²³ g
Solution:
Step 1) Calculate Molar Mass of SH₂;
Atomic Mass of Sulfur = 32 g/mol
Atomic Mass of H₂ = 2 g/mol
--------------------
Molecular Mass of SH₂ = 34 g/mol
Step 2: Calculate mass of one molecule of SH₂ as;
As,
Moles = # of Molecules / 6.022 × 10²³
Also, Moles = Mass / M.Mass So,
Mass/M.mass = # of Molecules / 6.022 × 10²³
Solving for Mass,
Mass = # of Molecules × M.mass / 6.022 × 10²³
Putting values,
Mass = (1 Molecule × 34 g.mol⁻¹) ÷ 6.022 × 10²³
Mass = 5.645 × 10⁻²³ g
<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X
<u>Explanation:</u>
To calculate the mole percent of a substance, we use the equation:
Mass percent means that the mass of a substance is present in 100 grams of mixture
To calculate the number of moles, we use the equation:
We require the molar masses of Z and X to calculate the mole percent of Z and X respectively
Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X
Answer:
A
Explanation:
it becomes a positive charge and fully filled
The reaction is;
2Na + 2H2O → 2NaOH + H2
Answer:
4.4 × 10^(23) atoms
Explanation:
The reaction is;
2Na + 2H2O → 2NaOH + H2
2 moles of Na produces 1 mole of H2 from the equation reaction.
Thus, since 8.2 L of hydrogen gas at STP are produced. And we know that at STP, 1 mole of a gas will have a volume of 22.4 L, and also, We know according to avogadro's number that, 1 mol = 6.02 × 10^(23) atoms, Thus;
8.2 L of H2 × (1 mole of H2/22.4 L) × (2 mole of Na/1 mole of H2) × (6.02 × 10^(23) atoms/1 mole of Na) = 4.4 × 10^(23) atoms
