2A + B + 2C ? D + E
1 answer:
Answer:
C Rate
Explanation:
In order to determine the correct rate law, let's use Trial 1 as baseline. Therefore:
An increase in [A] in Trial 2 by a factor of leads to an increase in the rate of reaction by a factor of 2 (i.e. the reaction rate is doubled). Thus, there is second order in [A].
Similarly,
An increase in [B] in Trial 3 by a factor of 1.667 leads to an increase in the rate of reaction by a factor of 1.667. Thus, there is first order in [B].
Futhermore,
An increase in [C] in Trial 4 by a factor of 1.71 leads to a decrease in the rate of reaction by 1.71. Thus, there is inverse first order in [C].
Therefore, the correct rate law is:
Rate
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Answer:
0.011 mol/L
Explanation:
This can be solved with something called an ICE table.
I = initial
C = change
E = equilibrium
Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.
x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.
After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.
Here it is in table form:
Now we can use the equilibrium constant:
Kc = [NO]² / ( [N₂] [O₂] )
Substituting:
0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )
Solving:
0.10 = (2x)² / (0.04 - x)²
√0.10 = 2x / (0.04 - x)
(√0.10) (0.04 - x) = 2x
(√0.10)(0.04) - (√0.10)x = 2x
(√0.10)(0.04) = 2x + (√0.10)x
(√0.10)(0.04) = (2 + √0.10)x
x = (√0.10)(0.04) / (2 + √0.10)
x = 0.0055
At equilibrium, the concentration of NO is 2x. So the answer is:
[NO] = 2x
[NO] = 0.011
The equilibrium concentration of NO is 0.011 mol/L.