The following mechanism for the gas phase reaction of Hy and ICI that is consist
ID: 1037429 • Letter: T
Question
The following mechanism for the gas phase reaction of Hy and ICI that is consistent with the observed rate law is: step 1 slow: H2() ICIgg HIg) step 2 fast. ICIg)+HIg)12) (1) What is the equation for the overall reaction? Use the smallestinteger coefficients posible If a box is not needed, leave it blank possible. If a box is not needed, leave it blank. (2) Which species acts as a catalyst? Enter formula. If none, leave box blank (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank 3 (4 Complete the rate law for the overall reaction that is consistent with this mechanism (Use the form k[A] [B1., where 'l' is understood (so don't write it) for m, n etc.) RateExplanation / Answer
Q1)
1) The overall reaction is
H2(g) + 2ICl(g) -------------> 2 Hcl(g) + I2(g)
2) No species acts as catalust
Thus NONE IS THE CATATLYST
3)The reaction intermediate is HI(g) [as it is formed in first step and consumed in the second step . It is not found in overall equation]
4)The rate law consistent with the mechanism is
rate = k [H2] [ICl]
The rate expression is always written from the slowest step of the reaction. As the slow step contains only one mole of H2 and ICl each the rate law is as given.
Q2)
1) The overall equation is
2NO(g) + Br2(g) -----------> 2 NOBr (g)
2)the species acting as catalyst is NONE
3) the reaction intermediate = NOBr2
4) The rate of the reaction is deduced from the slow step .
Thus
Rate = k [NOBr2][NO]
However the intermediate's concentration can not be calculated, thus using the first equilibirum, we can remove that term as
No + Br2 <----> NOBr2
Keq = [NOBr2] /[NO][Br2]
Thus [NOBr2] = Keq x [NO][Br2]
substituting this in the rate expression
Rate = k x Keq [NO][Br2] [NO]
So overall rate expression is
rate = k [No]2[Br2]
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