For the reaction:
`2NO_(2)(g) + F_(2)(g) to NO_(2)F +F` (slow)
ii) `NO_(2) + F to NO_(2)F + F` (fast)
What is the predicted rate law?
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`2A+2B to D +E ` for the reaction following mechanism has been proposed . `A+2B to 2C +d `( slow) ` A+2C to E ` ( fast ) The rate law expression for t
Correct Answer - C Rate of reaction for slowest step .
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The reaction `2A +B to B to D + E ` involves the mechanism , ` A to B `(fast) `B to C `(slow) ` A+C to D +E ` the rate expressio would be ,
Correct Answer - B Slowest step is the rate determining step , hence rate =k[B]
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The formation of `SO_(3) ` from ` SO_(2) and O_(2) ` takes place in the following steps : `(i) SO_(2)+2NO_(2) to 2SO_(3)+2NO` `(ii) 2NO +O_(2) to 2NO_
Correct Answer - C The reactants first combine with the catalyst to form an intermediate complex which is short lived and decomposes to form the product and regenerating the catalyst hence...
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Consider the reaction ` 2NO_((g) ) +O_(2(g) ) to 2NO_(2(g)).` if ` (d [NO_(2)])/(dt)=0.052 M//s ` then `-(d[O_(2)])/(dt)` will be
Correct Answer - C `"Rate"=-(1)/(2)(d[NO])/(dt)=-(d[O_(2)])/(dt)=+(1)/(2) (d[NO_(2)])/(dt)` `-(d[O_(2)])/(dt)=(1)/(2)xx0.052 =0.026 m//s`
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For the reversible reaction, net rate is `2NO(g)+O_(2)(g) hArr 2NO_(2)(g)` `((dx)/(dt))_("net")=2.6 xx 10^(3)[NO]^(2)[O_(2)]-4.1 [NO_(2)]^(2)` If a re
Correct Answer - B `K_(c )=(k_(f))/(k_(b))=(2.6 xx 10^(3))/(4.1)=6.34 xx 10^(2)` `Q_(c)=([NO_(2)]^(2))/([NO]^(2)xx[O_(2)])` `=([0.1 xx 1]^(2))/([0.01]^(2)[0.01])=10^(4)` i.e., `Q_(c )gt K` Hence the reaction shifts in backward direction so that reaction quotient `(Q_(c ))`...
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Nitrogen dioxide `(NO_(2))` reacts with fluroine `(F_(2))` to form nitryl fluroide `(NO_(2)F)` according to the reaction. `2NO_(2)(g) + F_(2) to 2NO_(
i) Rate of formation of `NO_(2)F1(r ) = (d(NO_(2)F))/(dt)` ii) Rate of disappearance of `NO_(2)`(r) = `(-d(NO_(2)))/(dt)` iii) Rate of disappearance of `F_(2)`(r) = `(-d(F_(2)))/(dt)`
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For a reaction carried at 400 K, `NO_(2)(g) + CO_(2)(g) to CO_(2)(g) + NO_(2)(g)`, the proposed mechanism is as follows: `NO_(2) + NO_(2) overset("Slo
The slow step for the reaction is: `NO_(2) + NO_(2) to NO + NO_(3)` (Slow) The rate law equation is: `Rate( r) = k[NO_(2)][NO_(2)]= k[NO_(2)]^(2)`.
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Writte the rate law and order for the following reaction: `AB_(2)+C_(2)to AB_(2)C+C` (slow) `AB_(2)+C to AB_(2)C` (fast)
The rate law for the reaction (r) = `k [AB_(2)][C_(2)]` Order of reaction = 1+1 = 2
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Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide `(NO_(2))` : `2NO(g) + O_(2) to 2NO_(2) (g)` The rate law for the reaction is: `rate
The possible mechanism for the reaction. `NO(g) + O_(2)(g) overset("slow")to NO_(2)(g)` `NO_(3)(g) + NO(g) overset("Fast")to 2NO_(2)(g)`
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Consider this reaction : `2NO_(2)(g)+O_(3)(g)rarrN_(2)O_(5)(g)+O_(2)(g)` The reaction of nitrogen diozidd and ozone represented is first order in `NO_
Correct Answer - C Mechanism I, rate =`k[NO_(2)][O_(2)]`. Slow step is rate determine step Mechanism II, Rate = `k[NO_(2)][O]" ".....(1)` `K=([O_(2)][O])/([O_(3)]),[O]=K([O_(3)])/([O_(2)])` From eq. (i) , Rate = `kK[NO_(2)[O_(3)][O_(2)]^(-1)` `:.` Both mechanisms...
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