A) 1
B) 0
C) 3
D) 2
Correct Answer: D
Solution :
Rate\[=k[NOB{{r}_{2}}][NO]\] ???..(i) But\[NOB{{r}_{2}}\]is in equilibrium. \[{{K}_{eq}}=\frac{[NOB{{r}_{2}}]}{[No][B{{r}_{2}}]}\] \[[NOB{{r}_{2}}]={{K}_{eq}}[NO][B{{r}_{2}}]\] ...(ii) Putting the value of\[[NOB{{r}_{2}}]\]in Eq. (i) Rate\[=k.{{K}_{eq}}[NO][B{{r}_{2}}][NO]\] Hence, rate\[=k.{{K}_{eq}}{{[NO]}^{2}}[B{{r}_{2}}]\] Rate\[=k'{{[NO]}^{2}}[B{{r}_{2}}]\] where \[k'=k.{{K}_{eq}}\] Alternative Solution \[NO(g)+B{{r}_{2}}(g)NOB{{r}_{2}}(g)\] \[NOB{{r}_{2}}(g)+NO(g)\xrightarrow{\,}NOBr(g)\] Rate\[=k[NOB{{r}_{2}}].[NO]=k.{{K}_{eq}}[NO][B{{r}_{2}}][NO]\] \[\left[ where,{{K}_{eq}}=\frac{[NOB{{r}_{2}}]}{[NO][B{{r}_{2}}]} \right]\] \[=k'{{[NO]}^{2}}[B{{r}_{2}}]\]You need to login to perform this action.
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