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Thepotemich [5.8K]
3 years ago
10

Consider the following reaction

Chemistry
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

92.72 kJ

Explanation:

2 N₂ (g) + O₂ (g) —-> 2 N₂O

According to question , one mole of N₂O requires 163.2 kJ of heat

Molecular weight of N₂O = 44 gm

25 g  N₂O = 25 / 44 mole

25 / 44 mole will require 163.2 x 25 / 44 kJ

= 92.72 kJ

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5 0
3 years ago
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
I need help with this for chemistry. I don’t understand now to do this.
alina1380 [7]

The ipR.O.B.O.T states

 aA+bB⇌ cC+dD  

the equilibrium constant is written as follows:

Kc=[C]c[D]d[A]a[B]b  

The ICE Table

The easiest approach for calculating equilibrium concentrations is to use an ICE Table, which is an organized method to track which quantities are known and which need to be calculated. ICE stands for:

"I" is for the "initial" concentration or the initial amount

"C" is for the "change" in concentration or change in the amount from the initial state to equilibrium

"E" is for the "equilibrium" concentration or amount and represents the expression for the amounts at equilibrium.

For the gaseous hydrogenation reaction below, what is the concentration for each substance at equilibrium?

C2H4(g)+H2(g)⇌C2H6(g)(1)

with  Kc=0.98  characterized from previous experiments and with the following initial concentrations:

[C2H4]0=0.33  

[H2]0=0.53  

SOLUTION

First the equilibrium expression is written for this reaction:

Kc=[C2H6][C2H4][H2]=0.98(2)

ICE Table

The concentrations for the reactants are added to the "Initial" row of the table. The initial amount of  C2H6  is not mentioned, so it is given a value of 0. This amount will change over the course of the reaction.

ICE

C2H4  

H2  

C2H6  

Initial

0.33

0.53

0

Change

Equilibrium

ICE

C2H4  

H2  

C2H6  

Initial

0.33

0.53

0

Change

-x

-x

+x

Equilibrium

Equilibrium is determined by adding "Initial" and "Change together.

ICE

C2H4  

H2  

C2H6  

Initial

0.33

0.53

0

Change

-x

-x

+x

Equilibrium

0.33-x

0.53-x

x

The expressions in the "Equilibrium" row are substituted into the equilibrium constant expression to find calculate the value of x. The equilibrium expression is simplified into a quadratic expression as shown:

0.98=x(0.33−x)(0.53−x)(3)

0.98=xx2−0.86x+0.1749(4)

0.98(x2−0.86x+0.1749)=x(5)

0.98x2−0.8428x+0.171402=x(6)

0.98x2−1.8428x+0.171402=0(7)

The quadratic formula can be used as follows to solve for x:

x=−b±b2−4ac−−−−−−−√2a(8)

x=−0.1572±(−0.1572)2−4(0.98)(0.171402)−−−−−−−−−−−−−−−−−−−−−−−−−√2(0.98)(9)

x=1.78 or0.098(10)

Because there are two possible solutions, each must be checked to determine which is the real solution. They are plugged into the expression in the "Equilibrium" row for  [C2H4]Eq :

[C2H4]Eq=(0.33−1.78)=−1.45(11)

[C2H4]Eq=(0.33−0.098)=0.23(12)

If  x=1.78  then  [C2H4]Eq  is negative, which is impossible, therefore,  x  must equal 0.098.

So:

[C2H4]Eq=0.23M(13)

[H2]Eq=(0.53−0.0981)=0.43M(14)

[C2H6]Eq=0.098M(15)

Problems

1. Find the concentration of iodine in the following reaction if the equilibrium constant is 3.76 X 103, and 2 mol of iodine are initially placed in a 2 L flask at 100 K.

I2(g)⇌2I−(aq)(16)

2. What is the concentration of silver ions in 1.00 L of solution with 0.020 mol of AgCl and 0.020 mol of Cl- in the following reaction? The equilibrium constant is 1.8 x 10-10.

AgCl(s)⇌Ag+(aq)+Cl−(aq)(17)

3. What are the equilibrium concentrations of the products and reactants for the following equilibrium reaction?

Initial concentrations:   [HSO−4]0=0.4   [H3O+]0=0.01   [SO2−4]0=0.07   K=.012  

HSO−4(aq)+H2O(l)⇌H3O+(aq)+SO2−4(aq)(18)

4. The initial concentration of HCO3 is 0.16 M in the following reaction. What is the H+ concentration at equilibrium? Kc=0.20.

H2CO3⇌H+(aq)+CO2−3(aq)(19)

5.The initial concentration of PCl5 is 0.200 moles per liter and there are no products in the system when the reaction starts. If the equilibrium constant is 0.030, calculate all the concentrations at equilibrium.

Solutions

1.

I2  

I−  

Initial

2mol/2L = 1 M

0

Change

−x  

+2x  

Equilibrium

1−x  

2x  

At equilibrium

Kc=[I−]2[I2]  

3.76×103=(2x)21−x=4x21−x  

cross multiply

4x2+3.76.103x−3.76×103=0  

apply the quadratic formula:

−b±b2−4ac−−−−−−−√2a  

with:  a=4 ,  b=3.76×103   c=−3.76×103 .

The formula gives solutions of of x=0.999 and -940. The latter solution is unphysical (a negative concentration). Therefore, x=0.999 at equilibrium.

[I−]=2x=1.99M(20)

[I2]=1−x=1−.999=0.001M(21)

2.

Ag+  

Cl−  

Initial

0

0.02mol/1.00 L = 0.02 M

Change

+x  

+x  

Equilibrium  

0.02+x  

Kc=[Ag−][Cl−](22)

1.8×10−10=(x)(0.02+x)(23)

x2+0.02x−1.8×1010=0(24)

x=9×10−9(25)

[Ag−]=x=9×10−9(26)

[Cl−]=0.02+x=0.020(27)

3.

H2CO3  

SO2−4  

H3O+  

Initial

0.4

0.01

0.07

Change

−x  

Equilibrium

0.4−x  

0.01+x  

0.07+x  

Kc=[SO2−4][H3O+]H2CO3(28)

0.012=(0.01+x)(0.07+x)0.4−x(29)

cross multiply and get:

x2+0.2x−0.0041=0(30)

apply the quadratic formula

x = 0.0328

[H2CO3]=0.4-x=0.4-0.0328=0.3672

[S042-]=0.01+x=0.01+0.0328=0.0428

[H30]=0.07+x=0.07+0.0328=0.1028

4.

H2CO3

H+  

CO2−3  

Initial

.16

0

Change

-x

Equilibrium

.16-x

apply the quadratic equation

x=0.1049

[H+]=x=0.1049

5. First write out the balanced equation:

PCl5(g)⇌PCl3(g)+Cl2(g)  

PCl5  

PCl3  

Cl2  

Initial

0.2

0

Change

-x

Equilibrium

0.2-x

Kc=[PC3][Cl2][PCl5](31)

0.30=x20.2−x(32)

Cross multiply:

x2+0.03x−0.006=0(33)

Apply the quadratic formula:

x=0.064

[PCl5]=0.2-x=0.136

[PCl3]=0.064

[Cl2]=0.064

Information is verified by Brainly Incorporations.

Do not copy this information without the consent of Brainly Inc.

ipR.O.B.O.T is an international Internet Protocol Recessive Observation Branch Organization Technologies

4 0
3 years ago
Which subatomic particle helps identify an elements identify
Sindrei [870]

Answer:

Protons

Explanation:

Protons are the subatomic particle that determines the identity of an element.

8 0
3 years ago
Serial dilution problem: Six test tubes are placed in a rack. To each tube add 4 mL of saline solution. Now to the first tube ad
ArbitrLikvidat [17]

Answer:

The dilution factor of protein in tube # 4 is 125. Molar concentration is 0.0088 M protein

Explanation:

The dilution factor indicates how many times is more concentrated a main solution in relationship with a diluted solution. In this case, the main solution is in tube #1. For calculating the dilution factor and molar concentration in tube #4 we need the main solution concentration which comes from next equation:

Initial volume * initial concentration = final volume * final concentration

0.5 mL * 10M = 5mL * final concentration

1.1 M = final concentration = main solution concentration

Applying the same equation for remain tubes we have 0.22 M for tube #2, 0.044 M for tube # 4 and 0.0088 for tube # 4.

Dilution factor = Main solution concentration/tube 4 concentration

Dilution factor = 1.1/0.0088 = 125

I hope my answer helps you

3 0
3 years ago
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