State collision theory
1 answer:
<h2>OR </h2>
Collision theory states that when suitable particles of the reactant hit each other with correct orientation, only a certain amount of collisions result in a perceptible or notable change; these successful changes are called successful collisions.
<h3>Explanation:</h3><h3>here your answer mate </h3>have a great day
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117 L. You can start by making a table to organize the information you are given. Then, you can use the formula PV/T=PV/T and plug in the numbers you have. You then solve for the missing volume. Remember that the initial pressure, temperature, and volume should be on one side of the equal sign, and the final pressure, volume, and temperature should be on the other side.
Answer:
The amount of NaF produced is doubled.
(d) is correct option.
Explanation:
Given that,
A 2 mole sample of F₂ reacts with excess NaOH according to the equation.
The balance equation is
If the reaction is repeated with excess NaOH but with 1 mole of F₂
The balance equation is
Hence, The amount of NaF produced is doubled.
(d) is correct option.
Answer:
a) [NOCl] = 0.968 M
[NO] = 0.032M
[Cl²] = 0.016M
b) [NOCl] = 1.992M
[NO] = 0.008 M
[Cl2] = 1.004 M
Explanation:
Step 1: Data given
Temperature = 35°C = 308K
K = 1.6 × 10^-5
Step 2: The reaction
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2
Step 3
a. 2.0 mol pure NOCl in a 2.0 L flask
Concentration at the start:
Concentration = mol / volume
[NOCl] = mol / volume
[NOCl] = 2.0 / 2.0 L
[NOCl] = 1.0 M
[NO] = 0 M
[Cl] = 0M
Concentration at the equillibrium
[NOCl] = 1.0M - 2x
[NO] = 2x
[Cl2]= x
K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5
1.6*10^-5 = ((2x)² * x) / (1.0-2x)²
x = 0.016
[NOCl] = 1.0 - 2*0.016 = 0.968 M
[NO] = 2*0.016 = 0.032M
[Cl²] = 0.016M
b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask
Concentration at the equillibrium
[NOCl] = 2.0 mol / 1.0 L = 2.0 M
[NO] = 0 M
[Cl2]= 1.0 mol / 1.0 L = 1.0 M
Concentration at the equillibrium
[NOCl] = 2.0M - 2x
[NO] = 2x
[Cl2]= 1.0 + x
K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5
1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)
1.6 *10^-5= (2x)² * 1 )/2.0²
1.6 *10^-5= 4x² / 4 = x²
x = = 4.0*10^-3
[NOCl] = 2.0 - 2*0.004 = 1.992M
[NO] = 2*0.004 = 0.008 M
[Cl2] = 1+ 0.004M = 1.004 M