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Hoochie [10]
3 years ago
14

The initial concentration of reactant in a first-order reaction is 0.27 M. The rate constant for the reaction is 0.75 s−1. What

is the concentration (mol/L) of reactant after 1.3 s? The initial concentration of reactant in a first-order reaction is 0.27 . The rate constant for the reaction is 0.75 . What is the concentration () of reactant after 1.3 ? 0.72 0.10 2.5 0.029 1.7
Chemistry
2 answers:
koban [17]3 years ago
5 0

Answer:

0.10 M

Explanation:

Given data

  • Initial concentration of the reactant ([A]₀): 0.27 M
  • Rate constant (k): 0.75 s⁻1
  • Time (t): 1.3 s
  • Concentration at time t ([A]): ?

For a first-order reaction, we can find the concentration of the reactant at a certain time using the following concentration vs. time expression.

[A]=[A]_{0}.e^{-k\times t}

[A]=0.27M.e^{-0.75s^{-1} \times 1.3s}

[A]=0.10M

e-lub [12.9K]3 years ago
4 0

Answer:

≈ 0.10M

Explanation:

[A]o = 0.27M

Rate constant, k = 0.75 s−1

[A] = ?

time, t = 1.3

Integrated formular for a first order reaction is given as;

ln[A] = ln[A]o − kt

ln[A] = ln(0.27) - 0.75(1.3)

ln[A] = -1.309 - 0.975

ln[A] = -2.284

[A] = 0.1019M ≈ 0.10M

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maks197457 [2]
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NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

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We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


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