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Chemistry 17 Online
OpenStudy (anonymous):

Could someone pleaseee explain this to me? I need help with (b) part, asking for the order of reaction.

OpenStudy (anonymous):

OpenStudy (wolfe8):

Would this help? http://www.chemguide.co.uk/physical/basicrates/orders.html

OpenStudy (anonymous):

I have read this before. I know how to calculate the order of reaction as well. What's confusing me is how do I calculate it with respect to NaOH when no values in the phenacyl chloride column are constant?

OpenStudy (anonymous):

You know, like for phenacyl chloride, comparing experiment 1 and 2 since NaOH is kept constant, the concenrtaion is increased 1.5 times and the rate also increases by 1.5 So, the order is 1 with respect to phenacyl chloride.

OpenStudy (wolfe8):

I think you can use the cross product of fractions. For example, \[\frac{ x }{ y }\times \frac{ y }{ z }=\frac{ x }{ z }\]

OpenStudy (anonymous):

Could you please elaborate on that a bit more? Like what values should I put in place of x, y and z?

OpenStudy (wolfe8):

You can have Rate of A/B times Rate of B/C so you'll end up with rate of A/C A/B can be the rate when one volume is constant. While rate of B/C can be the rate when another volume is constant. Sorry if I'm not making any sense. My mind is kinda loopy right now.

OpenStudy (anonymous):

well we know that in respects to the phenacyl c hloride n=1. n representing the exponent in the equation rate=k[phenacyl chloride]^n*[Naoh}^x. X is what you r solving for in this question so you just plug a data in and find your X by the laws of logarithms.

OpenStudy (anonymous):

give me more time on finding the numerical value but i believe this is the most correct way to solving for what your looking for. the order reaction rate laws.

OpenStudy (anonymous):

But without the value of rate constant k, you cannot get value of X.

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