NO reacts with H_2 in the gas phase according to the following chemical equation
ID: 482750 • Letter: N
Question
NO reacts with H_2 in the gas phase according to the following chemical equation: 2NO(g) +2H_2 (g) rightarrow N_2 (g) +2H-2 O(g) It is observed that, when the concentration of H_2 is reduced to 1/2 of its initial value, the rate of the reaction is also reduced to 1/2 of its initial value. When the concentration of NO is multiplied by 2.77, the rate of the reaction increases by a factor of 7.67. Write the rate expression for this reaction, and give the units of the rate constant k, assuming concentration is expressed as mol L^-1 and time is in seconds. If [NO] were multiplied by 2.86 and [H_2] by 2.22, what change in the rate would be observed? The rate would increase by a factor of .Explanation / Answer
(a) 2 NO (g) + 2 H2 (g) --------> N2 (g) + 2 H2O (g)
Let the rate law for the reaction be
R = k[NO]m[H2]n where m and n are the orders of the reaction w.r.t NO and H2.
Let [NO] = C1 and [H2] = C2; let R = R1.
Given R1 = k.C1m.C2n …..(1)
R2 = k.C1m.C2’n where C2’ = C2/2. Also, R2 = R1/2
Therefore,
R1/2 = k.C1m.(C1/2)n ….(2)
Divide (1) by (2) and get
R1/(R1/2) = C2n/(C2/2)n = 2n
====> 2 = 2n
====> n = 1
Again, R3 = k.C1’m.C2n
It is given that C1’ = 2.77*C1 and R3 = 7.67R1
Therefore,
7.67R1 = k.(2.77*C1)m.C2n ….(3)
Divide (3) by (1) and write
7.67 = (2.77*C1)m/(C1m) = (2.77)m
===> (2.77)2 = (2.77)m
===> m = 1
The rate law is Rate = k[NO]2[H2] (ans).
The rate of the reaction can be written as the rate of appearance of N2.
Rate of reaction = d[N2]/dt
Since [N2] is expressed in mol/L and t in sec, the unit for the reaction rate is
Unit of rate = (mol/L)/(s) = mol L-1 s-1.
Thus,
(Unit of Rate) = (unit of rate constant)*(unit of [NO])2*(unit of [H2])
===> (mol L-1 s-1) = (unit of k)*(mol/L)2.(mol/L)
===> mol L-1 s-1 = (unit of k)* mol3 L-3
===> unit of k = (mol L-1 s-1)/(mol3 L-3) = L2 mol-2 s-1 (ans).
(b) We have C1” = 2.86*C1 and C2” = 2.22*C2.
R3 = k.C1”2.C2” = k.(2.86*C1)2.(2.22*C2) = 18.158*k*C12*C2 18*R1
The reaction rate increases by 18 times (ans).
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