when I add $\ce{RCl + H2O -> ROH + HCl}$
but it's also true in reverse $\ce{ROH + HCl -> RCl +H2O}$
however this is not in case of $\ce{FeSO4 +Zn -> Zn + FeSO4}$
$\ce{ZnSO4 + Fe -> no rxn}$
How can a reaction be true both ways ?
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$\begingroup$ We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework. They are not banned, but we require a significant a priori effort. Otherwise, such a question may get closed. Please edit in your full reasoning or thoughts on this. See Homework $\endgroup$– PoutnikCommented Jun 1, 2019 at 13:21
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$\begingroup$ my mistake i will correct it $\endgroup$– BhavayCommented Jun 1, 2019 at 13:53
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3$\begingroup$ What does FeSO4 + Zn have anything to do with nucleophilic substitution? $\endgroup$– orthocresolCommented Jun 1, 2019 at 13:56
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$\begingroup$ If you insist on plain text, what is discouraged in Chemistry SE site, at least write the equations properly. Asking the question the way it is rather presents disrespect to readers. $\endgroup$– PoutnikCommented Jun 1, 2019 at 13:57
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1 Answer
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It should be better to write equations as
$$\ce{RCl +H2O <=> ROH +HCl}$$
$$\ce{Fe^2+ +Zn <=>> Zn^2+ + Fe }$$
with the latter having the equilibrium shifted to far right.
But is case of high concentration of $\ce{Fe^2+}$ and very low concentration of $\ce{Zn^2+}$, the net reaction would be ongoing leftwards.
If the left side of equation below is smaller than right one, the net reaction in inclined toward iron being replaced by zinc. And vice versa.
$$E^{\circ}_{\ce{Zn/Zn^2+}}+0.0295\log {a_{\ce{Zn^2+}}}=E^{\circ}_{\ce{Fe/Fe^2+}}+0.0295\log {a_{\ce{Fe^2+}}}$$
Note that forward and backward reaction are ongoing simultaneously, having the same rate at the equilibrium.
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$\begingroup$ what should be done to carry rxn in foward reaction $\endgroup$– BhavayCommented Jun 1, 2019 at 13:04