13In Q2) Take it 10-4 fr the rates....I don't have the courage to edit it now [130]
1) Assertion-Reason
I- Decomposition of NH3 on Pt surface is a zero order reaction at lower concentration of NH3, but at higher concentration, the order becomes a first order reaction.
II- Rate = \frac{K_1 \left[NH_3 \right]}{1+K_2 \left[NH_3 \right]}
a)Both correct;II explains I
b)Both correct;II dosen't explain I
c)I-True; II-False
d)I-False; II-True
2)For a given reaction, A--->P, Rate= 1x10-4 M/s when [A]=.01M & 1.41x 10-4 M/s when [A]=.02M.
Deduce the expression fr Rate Law.
3)The given reaction is-
xA-----> yB
& log_{10} \frac{-d[A]}{dt} = log_{10} \frac{d[B]}{dt} + 0.30
(-ve sign indicates rate of dissapearance of the reactant).
Find the value of- \frac{x^2 + y^2}{xy}
(P.S. 1st one is a doubt; Rest are not [1])
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9 Answers
13
106Q1.. think (c) .. (Both statements are contradictory.. so saying c)
13d) is the answer given... btw wht does the 2nd statement actually infer ??
1yes the answer is d because even at high concentrations the reaction remains a zero order case
29well in Qn. 1 the two statements given are contradictory so now we the options can be c or d..but as given in NCERT Pg 104 that decomposition of NH3 follows zeor order reaction at high conc. of NH3 so statement 1 is false ..hence the answer is d..
Qn 2>>>rate= k √A where k= 10-3
Qn 3>>> the answer is 5/2
1357Q1)
The second statement means that
when [NH3] is very low, K2[NH3] can be neglected w.r.t. 1
when [NH3] is high, 1 can be neglected w.r.t. K2[NH3]