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Tcecarenko [31]
3 years ago
5

Consider this initial-rate data at a certain temperature for the reaction described by N2O3(

Chemistry
2 answers:
Yuliya22 [10]3 years ago
6 0

<span><span>N2</span><span>O3</span><span>(g)</span>→NO<span>(g)</span>+<span>NO2</span><span>(g)</span></span>

<span><span>[<span>N2</span><span>O3</span>]</span> Initial Rate</span>
<span>0.1 M     r<span>(t)</span>=0.66</span> M/s
<span>0.2 M     r<span>(t)</span>=1.32</span> M/s
<span>0.3 M     r<span>(t)</span>=1.98</span> M/s

We can have the relationship:

<span>(<span><span>[<span>N2</span><span>O3</span>]/</span><span><span>[<span>N2</span><span>O3</span>]</span>0</span></span>)^m</span>=<span><span>r<span>(t)/</span></span><span><span>r0</span><span>(t)
However,
</span></span></span>([N2O3]/[N2O3]0) = 2

Also, we assume m=1 which is the order of the reaction.

Thus, the relationship is simplified to,

r(t)/r0(t) = 2

r<span>(t)</span>=k<span>[<span>N2</span><span>O3</span>]</span>

0.66 <span>M/s=k×0.1 M</span>

<span>k=6.6</span> <span>s<span>−<span>1</span></span></span>

34kurt3 years ago
3 0

The value of k is \boxed{\text{6.6 s}^{-1}} .

Further Explanation:

Order of reaction:

It is sum of exponents of concentration of each of the reactants that are present in chemical reaction. In other words, order indicates power dependence of order of reaction on reactant concentration.

Given reaction is as follows:

 \text{N}_2\text{O}_3(\text{g})\rightarrow\text{NO}(\text{g})+\text{NO}_2(\text{g})

When concentration of  \text{N}_2\text{O}_3 changes from 0.1 M to 0.2 M, rate of reaction becomes doubled. This implies order of reaction with respect to  \text{N}_2\text{O}_3 is one.

The expression for rate of given reaction is as follows:

\text{Rate}=\text{k}[\text{N}_2\text{O}_3]                                         ...... (1)

Where, k is the rate constant of reaction.

Rearrange equation (1) to calculate the value of k.

\text{k}=\dfrac{\text{Rate}}{[\text{N}_2\text{O}_3]}                                                     ...... (2)

Any of the three given concentrations of  \text{N}_2\text{O}_3 can be considered for calculation of rate constant along with the respective initial rates of reaction.

Here, we take concentration of \text{N}_2\text{O}_3 as 0.1 M so rate corresponding to this is 0.66 M/s.

Substitute 0.1 M for  [\text{N}_2\text{O}_3] and 0.66 M/s for rate in equation (2).

 \begin{aligned}{\text{k}&=\dfrac{\text{0.66 M/s}}{\text{0.1 M}}\\&=\text{6.6 s}^{-1}}\end{aligned}

 

Therefore the value of rate constant (k) for the given reaction comes out to be \text{6.6 s}^{-1}}.

Learn More:  

1. What is the half-life of the reaction? brainly.com/question/8907464  2. Rate of chemical reaction: brainly.com/question/1569924  

Answer Details:  

Grade: Senior School  

Subject: Chemistry  

Chapter: Chemical Kinetics  

Keywords: k, rate, order, N2O3, NO, NO2, 0.1 M, 0.66, 0.2 M, 1.32, 0.3 M, 1.98, rate constant, sum, exponents.

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3 years ago
The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
3 years ago
A hot air balloon is filled with 15 moles helium gas and 5 moles nitrogen gas. What is the volume of the balloon at 1.01 atm and
kvasek [131]

<u>Given:</u>

Moles of He = 15

Moles of N2 = 5

Pressure (P) = 1.01 atm

Temperature (T) = 300 K

<u>To determine:</u>

The volume (V) of the balloon

<u>Explanation:</u>

From the ideal gas law:

PV = nRT

where P = pressure of the gas

V = volume

n = number of moles of the gas

T = temperature

R = gas constant = 0.0821 L-atm/mol-K

In this case we have:-

n(total) = 15 + 5 = 20 moles

P = 1.01 atm and T = 300K

V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L

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3 0
3 years ago
How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?
allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

 

First let us calculate the number of moles of AgNO3.

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moles AgNO3 = 0.011025 mol

 

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

 

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

<span>volume Na2S = 0.0212 L = 21.2 mL</span>

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Mandarinka [93]

Answer:

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Explanation:

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4 0
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