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The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment in a CSTR. When pure A is fed to a 10 dm 3 PFR at 300 K and a volumetric flow rate of 5 dm 3 /s, the conversion is 80%. When a mixture of 50% A and 50% inert (I) is fed to a 10 dm 3 CSTR at 320 K and a volumetric flow rate of 5 dm 3 /s, the conversion is also 80%. What is the activation energy in cal/mol

Answers

Answer 1

Answer:

Answer:

The activation energy is $=8.1\,kcal\,mol^(-1)$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_(A)=kC_(A)$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^(3)$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_(o)=5dm^(3)s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= (v_(0))/(k)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]$

Rearrange the formula is as follows.

$k= (v_(0))/(V)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_(o)}$ is 1.

$\epsilon =y_{A_(o)}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=(5m^(3)/s)/(10dm^(3))[(1+1)ln (1)/(1-0.8)-1 * 0.8] = 1.2s^(-1)$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^(-1)$

The rate constant in case of the CSTR can be calculated by using the formula.

$(V)/(v_(0))= (X(1+\epsilon X))/(k(1-X)).............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_(o)}$ is 0.5

$\epsilon =y_{A_(o)}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$(10dm^(3))/(5dm^(3))=(0.8(1+0.5(0.8)))/(k(1-0.8))=2.8s^(-1)$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^(-1)$

The activation energy of the reaction can be calculated by using formula

$k(T_(2))=k(T_(1))exp[(E)/(R)((1)/(T_(1))-(1)/(T_(2)))]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R *((T_(1)T_(2))/(T_(1)-T_(2)))ln(k(T_(2)))/(k(T_(1)))$

Substitute the all values.

$=1.987cal/molK((300K *320K)/(320K *300K))ln (2.8)/(1.2)=8.081 *10^(3)cal\,mol^(-1)$

$=8.1\,kcal\,mol^(-1)$

Therefore, the activation energy is $=8.1\,kcal\,mol^(-1)$

2.57 × 10²⁴ molecules

Explanation:

The number of molecules can be found by using the formula

N = n × L

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 4.27 × 6.02 × 10²³

We have the final answer as

2.57 × 10²⁴ molecules

Hope this helps you

How many electrons are in Fe3+ ?

Answers

23 electrons hope this helps

Answer:

There are 23 electrons in Fe3+

When palmitoleic acid reacts with hydrogen to form a saturated fatty acid, indicate the stoichiometry of the reaction and the product that is formed. If the stoichiometry of H2 or the product is not integral, enter a fraction (i.e. 3/2)

Answers

Answer:

Stoichiometric coefficient of hydrogen gas is 1.

Stoichiometric coefficient of palmitic acid is 1.

Explanation:

Addition of hydrogen to double bond is termed as hydrogenation reaction.

$C_(16)H_(30)O_2+H_2\rightarrow C_(16)H_(32)O_2$

According to stoichiometry, 1 mole of palmitoleic acid reacts with 1 mole of hydrogen gas to give 1 mole of palmitic acid.

Stoichiometric coefficient of hydrogen gas is 1.

Stoichiometric coefficient of palmitic acid is 1.

Calculate the volume of 0.100 m hcl required to neutralize 1.00 g of ba(oh)2 (molar mass = 171.3 g/mol).

Answers

The balanced chemical equation between HCl and $Ba(OH)_(2)$ is:

$2HCl (aq) + Ba(OH)_(2)(aq) -->BaCl_(2)(aq) + 2 H_(2)O(l)$

Moles of $Ba(OH)_(2)$ = $1.00 g Ba(OH)_(2) * (1 mol Ba(OH)_(2))/(171.3 g Ba(OH)_(2)) = 0.00584 mol Ba(OH)_(2)$

Moles of HCl required to neutralize $Ba(OH)_(2)$ :

$0.00584 mol Ba(OH)_(2) * ( 2 mol HCl)/(1 mol Ba(OH)_(2)) = 0.01168 mol HCl$

Calculating the volume of HCl from moles and molarity:

$0.01168 mol HCl * (1 L)/(0.100 mol) * (1000 mL)/(1 L) = 116.8 mL$

Answer:- 117 mL of HCl are used.

Solution:- The balanced equation for the reaction of HCl with barium hydroxide is written as:

$2HCl(aq)+Ba(OH)_2(aq)\rightarrow BaCl_2(aq)+2H_2O(l)$

From above equation, HCl and $Ba(OH)_2$ react in 2:1 mol ratio.

We will calculate the moles of barium hydroxide on dividing its grams by its molar mass.

Molar mass of Barium hydroxide is given as 171.3 g per mol.

$1.00gBa(OH)_2((1mol)/(171.3g))$

= 0.00584 mol $Ba(OH)_2$

Using mol ratio we calculate the moles of HCl as:

$0.00584molBa(OH)_2((2molHCl)/(1molBa(OH)_2))$

= 0.01168 mol HCl

We know that molarity is moles of solute per liter of solution. We have 0.01168 moles of HCl and its molarity is 0.100 M. So, we can calculate the liters of HCl solution used on dividing the moles by molarity as and on multiplying by 1000 the liters are converted to mL since, 1 L = 1000 mL.

$0.01168mol((1L)/(0.100mol))((1000mL)/(1L))$

= 116.8 mL

It could be round to 117 mL.

So, 117 mL of HCl are required.

The solubility of glucose at 30°C is125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.

Answers

Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

$400 mL * (0.996g)/(1mL) =398g$

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

$100gH_2O * (550gGlucose)/(398gH_2O) = 138 gGlucose$

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated. If you want to increase the amount of glucose in the solution without adding more glucose, you can increase the temperature.

The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated.

Since the solubility of glucose at 30°C is 125 g/100 g water, adding 550 g of glucose to 400 mL of water exceeds the maximum amount of glucose that can dissolve in the given amount of water.

To increase the amount of glucose in the solution without adding more glucose, you can increase the temperature. Higher temperatures generally increase the solubility of solutes in water. By increasing the temperature, you can dissolve more glucose in the solution.

Learn more about solubility here:

brainly.com/question/31493083

#SPJ3

Pls help ASAP I will give brainliest

Answers

Answer:

Lemon

HCI

Blood

Saliva

Bleach

NaOH

Explanation:

Blood 7.35-7.45

Bleach 12.6

Saliva 6.2-7.6

Lemon 2-3

HCI 3.01

NaOH 13

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