Rate help please!!
C2H4(g) + H2(g) ----> C2H6(g)
Rate = k[C2H4]
At a fixed temp, the reaction mixture is compressed to triple the original pressure.
What is the factor by which the rate of reaction changes?
A) 6
B) 9
C) 12
D)27
Confused as to how to approach this question?
Rate = k[C2H4]
At a fixed temp, the reaction mixture is compressed to triple the original pressure.
What is the factor by which the rate of reaction changes?
A) 6
B) 9
C) 12
D)27
Confused as to how to approach this question?
(edited 4 years ago)
Original post by anactualmess
C2H4(g) + H2(g) ----> C2H6(g)
Rate = k[C2H4]
At a fixed temp, the reaction mixture is compressed to triple the original pressure.
What is the factor by which the rate of reaction changes?
Confused as to how to approach this question?
Rate = k[C2H4]
At a fixed temp, the reaction mixture is compressed to triple the original pressure.
What is the factor by which the rate of reaction changes?
Confused as to how to approach this question?
Under the condition the temperature is not only fixed, but also constant:
a higher pressure causes that the volume decreases that in turn means the reactants have lesser space to move around and thus the rate of collision between the molecules/particles increases. And that leads to a higher likelihood they react each other. In conclusion to these thoughts, the rate must increased when the pressure is tripled.
(edited 4 years ago)
Thank you, but I forgot to add the rest of the question, I've edited the question now, it is these options which confuse me as I wouldn't know the exact factor but I would know that rate does increase.
Original post by Kallisto
Under the condition the temperature is not only fixed, but also constant:
a higher pressure causes that the volume decreases that in turn means the reactants have lesser space to move around and thus the rate of collision between the molecules/particles increases. And that leads to a higher likelihood they react each other. In conclusion to these thoughts, the rate k must increased when the pressure is tripled.
a higher pressure causes that the volume decreases that in turn means the reactants have lesser space to move around and thus the rate of collision between the molecules/particles increases. And that leads to a higher likelihood they react each other. In conclusion to these thoughts, the rate k must increased when the pressure is tripled.
Original post by anactualmess
C2H4(g) + H2(g) ----> C2H6(g)
Rate = k[C2H4]
At a fixed temp, the reaction mixture is compressed to triple the original pressure.
What is the factor by which the rate of reaction changes?
Confused as to how to approach this question?
Rate = k[C2H4]
At a fixed temp, the reaction mixture is compressed to triple the original pressure.
What is the factor by which the rate of reaction changes?
Confused as to how to approach this question?
Original post by Kallisto
Under the condition the temperature is not only fixed, but also constant:
a higher pressure causes that the volume decreases that in turn means the reactants have lesser space to move around and thus the rate of collision between the molecules/particles increases. And that leads to a higher likelihood they react each other. In conclusion to these thoughts, the rate k must increased when the pressure is tripled.
a higher pressure causes that the volume decreases that in turn means the reactants have lesser space to move around and thus the rate of collision between the molecules/particles increases. And that leads to a higher likelihood they react each other. In conclusion to these thoughts, the rate k must increased when the pressure is tripled.
Kallisto, you're right that the rate increases, but k stays the same. Increasing the pressure of a gas is the same as decreasing the volume by the same factor, as you point out, but crucially, this is also the same as increasing the gases' concentrations by the same factor. So if the pressure is tripled, both [C2H4] and [H2] also triple. You have a formula for working out how the rate changes when those concentrations triple. I hope this helps.
thank you so much!
(Original post by anosmianAcrimony)
(Original post by anosmianAcrimony)
Kallisto, you're right that the rate increases, but k stays the same. Increasing the pressure of a gas is the same as decreasing the volume by the same factor, as you point out, but crucially, this is also the same as increasing the gases' concentrations by the same factor. So if the pressure is tripled, both [C2H4] and [H2] also triple. You have a formula for working out how the rate changes when those concentrations triple. I hope this helps.
Original post by anactualmess
thank you so much!
You are welcome! For a more in-depth explanation:
https://www.chemguide.co.uk/physical/basicrates/pressure.html
Original post by anosmianAcrimony
Kallisto, you're right that the rate increases, but k stays the same. Increasing the pressure of a gas is the same as decreasing the volume by the same factor, as you point out, but crucially, this is also the same as increasing the gases' concentrations by the same factor. So if the pressure is tripled, both [C2H4] and [H2] also triple. You have a formula for working out how the rate changes when those concentrations triple. I hope this helps.
Thanks that you discovered my little mistake, corrected it. Very kind of you that you told me.
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