What would happen to the rate of a reaction with rate law rate $=k[ NO ]^2\[ H _2\]$ if the concentration of NO were halved?
A. The rate would be doubled.
B. The rate would be four times larger.
C. The rate would be one-fourth.
D. The rate would also be halved.
Answer (1)
The initial rate is r a t e 1 = k [ NO ] 2 [ H 2 ] .
The new concentration of NO is [ NO ] n e w = 2 [ NO ] .
The new rate is r a t e 2 = k ( 2 [ NO ] ) 2 [ H 2 ] = 4 1 k [ NO ] 2 [ H 2 ] .
The new rate is one-fourth of the original rate: C .
Explanation
Problem Analysis Let's analyze the problem. We are given a rate law: r a t e = k [ NO ] 2 [ H 2 ] . We need to determine how the rate changes when the concentration of NO is halved.
Setting up the Equations Let's denote the initial rate as r a t e 1 = k [ NO ] 2 [ H 2 ] . Now, let's halve the concentration of NO , so the new concentration is [ NO ] n e w = 2 [ NO ] .
Calculating the New Rate The new rate, r a t e 2 , can be expressed as:
r a t e 2 = k [ NO ] n e w 2 [ H 2 ] = k ( 2 [ NO ] ) 2 [ H 2 ] = k ( 4 [ NO ] 2 ) [ H 2 ] = 4 1 k [ NO ] 2 [ H 2 ]
Comparing Rates Now, let's compare the new rate ( r a t e 2 ) to the initial rate ( r a t e 1 ):
r a t e 1 r a t e 2 = k [ NO ] 2 [ H 2 ] 4 1 k [ NO ] 2 [ H 2 ] = 4 1
Final Answer This shows that the new rate is one-fourth of the original rate. Therefore, if the concentration of NO is halved, the rate becomes one-fourth of the original rate. The correct answer is C.
Examples
Understanding reaction rates is crucial in many real-world applications. For example, in the Haber-Bosch process, which produces ammonia for fertilizers, controlling the concentrations of nitrogen and hydrogen is essential to optimize the production rate. If the concentration of nitrogen were halved, the production rate would decrease significantly, impacting the efficiency of fertilizer production. Similarly, in pharmaceutical manufacturing, precise control of reactant concentrations ensures consistent drug synthesis rates, maintaining product quality and efficacy.
