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Crank
3 years ago
8

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

rroundings
Chemistry
1 answer:
anastassius [24]3 years ago
8 0

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

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The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0
3 years ago
I need help on thiss
Sergio039 [100]

The reaction is not balanced

<h3>Further explanation</h3>

Given

Reaction

2Fe(s)+3O₂(g)⇒2Fe₂O₃(s)

Required

The number of atoms

Solution

In a balanced chemical equation, the number of atoms in the compound that reacts (the reactants and products) will have the same number

Reactants : Fe(s)+O₂(g)

Fe = 2 atoms

O = 3 x 2 = 6 atoms

Products : Fe₂O₃(s)

Fe = 2 x 2 = 4 atoms

O = 2 x 3 = 6 atoms

The reaction is not balanced because the number of Fe atoms is not the same

The balanced reaction should be:

4Fe(s)+3O₂(g)⇒2Fe₂O₃(s)

8 0
2 years ago
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