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3 years ago

The rate of a reaction or how quickly a reaction will proceed is best determined by?

Chemistry

1 answer:

lilavasa [31]3 years ago

3 0

Answer:

Gibbs Free Energy (ΔG)

Explanation:

Gibbs Free Energy (G) - Is the energy associated with a chemical reaction that can be used to do useful work. The free energy of a system is the sum of its enthalpy (H) minus the product of the temperature (Kelvin) and the entropy (S) of the system:

G = H -TS--------------------------------------------------------- (1)

ΔG = ΔH - TΔS----------------------------------------------(2)

ΔG = Negative, means the reaction is spontaneous

ΔG = Positive, means the reaction is slow

ΔG = Zero, means the reaction is in equilibrium

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4.True or false. P(E) + P(Ec) = 1.

iogann1982 [59]

4.) True:
Since
P(E) = 1 - P(Ec)
P(E)+P(Ec)=1
If you think of it as the chances of you getting e or not getting e, then it's a 100% chance of either one happening.

4 0

3 years ago

A compound with a molar mass of 60g/mol is 40.4% carbon, 6.7% hydrogen and 53.3% oxygen (by mass). determine the emperical and m

Fittoniya [83]

<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.

C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1

Empirical formula - CH₂O

Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂

4 0

2 years ago

A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc

Paul [167]

Answer : The volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of $O_2$ gas = (740-22.4) torr = 717.6 torr

$P_2$ = final pressure of $O_2$ gas at STP= 760 torr

$V_1$ = initial volume of $O_2$ gas = 280 mL

$V_2$ = final volume of $O_2$ gas at STP = ?

$T_1$ = initial temperature of $O_2$ gas = $25^oC=273+25=298K$

$T_2$ = final temperature of $O_2$ gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get:

$\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}$

$V_2=242.2mL=0.2422L$

Therefore, the volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

5 0

3 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

2 years ago

A molecule of carbon dioxide is made up of 1 atom of carbon and 2 atoms of oxygen. Which of the following represents 2 carbon di

Degger [83]

Here’s a CO2 molecule :)

3 0

2 years ago

Read 2 more answers