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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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Isobutyl propionate is the substance that provides the flavor for rum extract. combustion of a 1.152 g sample of this carbon-hyd
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Answer;

C7H14O2

Solution;

Isobutyl contains , oxygen, carbon and hydrogen (total mass is 1.152 g)

Mass of carbon = 12/44 × 2.726 g

                           = 0.743455 g

Mass of Hydrogen  = 2/18 × 1.116 g

                            =  0.124 g

Mass of oxygen = 1.152 - (0.7435 + 0.124)

                           =  0.2845 g

Moles of carbon ;  0.7435/12 = 0.06196 moles

Moles of hydrogen; 0.124/1 = 0.124 moles

Moles of oxygen;  0.2845/16 = 0.01778 moles

Ratios ; 0.06196/0.01778 ; 0.124/0.01778 : 0.01778/0.01778

         =  3.5 :  7.0 : 1

To make them whole numbers ; we multiply the ratios by 2  to get;

(3.5 :  7.0 : 1 )2 = 7 : 14 : 2

Thus, the empirical formula of Isobutyl propionate  is C7H14O2


4 0
3 years ago
Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (C2H5OH) using the standard enthalpies of fo
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Answer:

Explanation:

For the reaction

C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O

We can calculate the  standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:

ΔHºc =  2ΔHºf CO2 (g) + 3ΔHºfH2O(l)  - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )

( we were not give the water state but we know we are at standard conditions so it is in its liquid state )

The ΔHºfs can be found in appropiate reference or texts.

ΔHºc =  2ΔHºf CO2 (g)+ 3ΔHºfH2O(l)  - (  ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )

= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [(  -276.2 + 0 ) ] kJ

ΔHºc =  -1368.33 kJ

5 0
3 years ago
What pressure (in atmospheres) does a gas exert when at 385 mm of Hg?
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2. How many moles are in 2.8 Liters of CO2 gas?
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Answer:

0.125 moles

Explanation:

2.8 litres is equivalent to 2.8dm³

At STP,

1 mole = 22.4 dm³

x mole = 2.8 dm³

Cross multiply

22.4x = 2.8

Divide both sides by 22.4

x = 2.8/22.4

x = 0.125

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2 years ago
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