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From the concetration of C_(4)H_(9)Cl (butyl chloride) at different times given below, calculate the average rate of reaction: C_(4)H_(9)Cl+H_(2)O rarr C_(4)H_(9)OH+HCl during different intervals of time. |{:([C_(4)H_(9)Cl] (mol L^(-1)),t (s)),(0.100,0),(0.0905,50),(0.0820,100),(0.0741,150),(0.0671,200),(0.0549,300),(0.0439,400),(0.0210,700),(0.017,800):}| |
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Answer» Solution :First determine the difference in the concentration over difference intervals of time and thus determine the average rate by dividing `Delta[R]` by `Delta t`. Average rate of hydrolyiss of butyl CHLORIDE  It is clear that the average rate falls form `1.90 xx 10^(-4) mol L^(-1)s^(-1)` to `0.4 xx 10^(-4) mol L^(-1) s^(-1)`. However, the average rate cannot be used to predict the rate of a reaction at a particular instant as it WOULD br constant for the time interval for which it is calculated. So to express the rate at a particular moment of time, the instantaneous rate is determine. It is obtained when we conisder the average rate at the smallest time interval say `dt` (i.e., when `Delta t` APPROACHES zero). Hence, mathematically, for an infinitesmally small `dt`, instantaneous rate is given by: `r_(av) = (-Delta[R])/(Delta t) = (Delta[P])/(Delta t)` As `t rarr 0, ot r_("inst") = (-d[R])/(dt) = (d[P])/(dt)`...(i) It can be determined graphically by drawing a tangent at time `t` on either isdes of the curves for concentration of `R` and `P` vs time `t` calculatingits slope. So in ILLUSTRATION 4.1, `r_(inst)` at `600 s`, for example, can be calculated by plotting the concentration of butyl chloride as a function of time. A tangent is drawn that touches the curve at `t = 600 s`.  The slope of this tangent gives the instantaneous rate. So,`r_("inst")` at `600 s = ((0.0165 - 0.037)/((800-400)s))mol L^(-1)` `= 5.12 xx 10^(-5) mol L^(-1) s^(-1)` At `t = 250s, r_("inst") = 1.22 xx 10^(-4) ,ol L^(-1) s^(-1)` `t = 350 s, r_("inst") = 1.0 xx 10^(-4) mol L^(-1) s^(-1)` `r = 450 s, r_("inst") = 6.4 xx 10^(-5) mol L^(-1) s^(-1)` |
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