The rate of disappearance of chlorine gas : 0.2 mol/dm³
<h3>Further explanation</h3>
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
For reaction :
The rate reaction :
![\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}](https://tex.z-dn.net/?f=%5Ctt%20-%5Cdfrac%7B1%7D%7Ba%7D%5Cdfrac%7Bd%5B-A%5D%7D%7Bdt%7D%3D%20-%5Cdfrac%7B1%7D%7Bb%7D%5Cdfrac%7Bd%5B-B%5D%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7Bc%7D%5Cdfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7Bd%7D%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D)
Reaction for formation CCl₄ :
<em>CH₄+4Cl₂⇒CCl₄+4HCl</em>
<em />
From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³
Rate of formation of CCl₄ = reaction rate x coefficient of CCCl₄
0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³
The rate of disappearance of chlorine gas (Cl₂) :
Rate of disappearance of Cl₂ = reaction rate x coefficient of Cl₂
Rate of disappearance of Cl₂ = 0.05 x 4 = 0.2 mol/dm³
The question is incomplete.
You need two additional data:
1) the original volume
2) what solution you added to change the volume.
This is a molarity problem, so remember molarity definition and formula:
M = n / V in liters: number of moles per liter of solution
To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).
The original solution was:
V= 1 ml
M = 0.2 M
Using the formula for molarity, M = n / V
n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles
For the final solution:
n = 0.0002 moles
M = 0.04
From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l
Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.
I hope this gives you all the cues you need to answer similar problems about molarity.
I calculated is and the answer is 1272.0600000000002.
I really don't know how to explain it
Physical, since you are physically changing its appearance
