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lina2011 [118]
3 years ago
9

For the reaction: 2 H2 + 2 NO → N2 + 2 H2O the observed rate expression, under some conditions, is: rate = k[H2][NO]2 Which of t

he following mechanisms are consistent with these data? Select all that are True. step 1 H2 + 2 NO → N2O + H2O (slow) step 2 N2O + H2 → N2 + H2O (fast) step 1 2 H2 + 2 NO → N2 + H2O step 1 NO + NO ⇌ N2O2 (fast) step 2 N2O2 + H2 → N2 + H2O + O (slow) step 3 O + H2 → H2O (fast) step 1 H2 + NO ⇌ H2ON (fast) step 2 H2ON + NO → N2 + H2O2 (slow) step 3 H2O2→ H2O + O (fast) step 4 O + H2 → H2O (fast)
Chemistry
1 answer:
Andrew [12]3 years ago
8 0

Answer:

Step 1 H2 + 2 NO → N2O + H2O (slow)

step 2 N2O + H2 → N2 + H2O (fast)

Explanation:

It is known that the slowest step in a reaction is the rate determining step in a sequence of reactions (reaction mechanism).

We have two important pieces of information in the question to guide our decision making process.

The overall reaction equation, and the rate expression. The two;

2 H2 + 2 NO → N2 + 2 H2O and rate = k[H2][NO]2 all support the answer given above.

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6 0
3 years ago
Use the "x is small" approximation to find the concentration of the products in the following reaction which initially contains
Mrrafil [7]

Answer:

Concentration of product at equilibrium ;

[H^+]=0.0000229 M

[CN^-]=0.0000229 M

Explanation:

HCN(aq)\rightleftharpoons H^+(aq) + CN^-(aq)

initially

0.85 M        0    0

(0.85-x)M    x      x

The equilibrium constant of reaction = K_c= 6.17\times 10^{-10}

The expression of an equilibrium cannot can be written as:

K_c=\frac{[H^+][CN^-]}{[HCN]}

6.17\times 10^{-10}=\frac{x\times x}{(0.85-x)}

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x = 0.0000229

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4 0
3 years ago
The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp
Eva8 [605]

Answer:

[A]_0=0.400M

Explanation:

Hello.

In this case, since the first-order reaction is said to be linearly related to the rate of reaction:

r=-k[A]

Whereas [A] is the concentration of hydrogen peroxide, when writing it as a differential equation we have:

\frac{d[A]}{dt} =-k[A]

Which integrated is:

ln(\frac{[A]}{[A]_0} )=-kt

And we can calculate the initial concentration of the hydrogen peroxide as follows:

[A]_0=\frac{[A]}{exp(-kt)}

Thus, for the given data, we obtain:

[A]_0=\frac{0.321M}{exp(-2.54x10^{-4}s^{-1}*855s)}

[A]_0=0.400M

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