Answer:
The rate of the reaction will increase by a factor of 9.
Explanation:
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In this case, considering the given second-order reaction, whose rate law results:
We easily infer that at constant concentration of A but tripling the concentration of B, we are going to obtain the following increasing factor while holding the remaining variables constant:
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Answer:
The heat of vaporisation of methanol is "3.48 KJ/Mol"
Explanation:
The amount of heat energy required to convert or transform 1 gram of liquid to vapour is called heat of vaporisation
When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.
Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is =
= 3.48 KJ
So, the heat of vaporization
Therefore, the heat of vaporization of methanol is 3.48KJ/Mol