InterviewSolution
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InterviewSolution
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The data given below are for the reaction of `NO` and `Cl_(2)` to from `NOCl` at `295 K`. `{:([Cl_(2)],[NO],,,"Initial rate" (mol litre^(-1) sec^(-1))),(0.05,0.05,,,1xx10^(-3)),(0.15,0.05,,,3xx10^(-3)),(0.05,0.15,,,9xx10^(-3)):}` (a) What is the order with respect to `NO` and `Cl_(2)` in the reaction? (b) Write the rate expression. (c ) Calculate the rate constant. (d) Determine the reaction rate when conc. of `Cl_(2)` and `NO` are `0.2 M` and `0.4 M` respectively. |
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Answer» For the reaction, `2NO+Cl_(2)rarr 2NOCl` Rate=`K[Cl_(2)]^(m)[NO]^(n) …(1)` Where, m and n are order of reaction w.r.t. `Cl_(2)` and `NO`, respectively. From the given data: `1xx10^(-3)= K[0.05]^(m)[0.05]^(n) …(2)` `3xx10^(-3)= K[0.15]^(m)[0.05]^(n) ...(3)` `9xx10^(-3)= K[0.05]^(m)[0.15]^(n) ...(4)` By eqs. (2) and (3), `m=1` By eqs. (2) and (4), `n=2` (a) `:.` Order w.r.t. `NO` is `2` and w.r.t. to `Cl_(2)` is `1`. (b) Also, rate expression `r=K[Cl_(2)]^(1) [NO]^(2)` (c ) And rate constant, `K=(r)/([Cl_(2)]^(1)[NO]^(2))` `=(1xx10^(-3))/([0.05]^(1)[0.05]^(2))` `=8 litre^(2) mol^(-2) sec^(-1)` (d) Further, `r=K[Cl_(2)]^(1)[NO]^(2)=8[0.2]^(1)[0.4]^(2)` `=0.256 mol litre^(-1) sec^(-1)` |
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