All categories

2 years ago

Consider the reaction 3CH4(g)→C3H8(g)+2H2(g).

Chemistry

1 answer:

padilas [110]2 years ago

6 0

From the calculations performed, the free energy change for the reaction is 72 kJ/mol.

<h3>What is the equilibrium constant?</h3>

The equilibrium constant is a value that shows the extent to which reactants have been converted to products.

Given that the equation of the reaction is;

3CH4(g)→C3H8(g)+2H2(g)

Then;

PC3H8 = 0.013 atm

PH2 = 2.3×10−2 atm

PCH4 = 41 atm

Now;

ΔG = ΔG° + RTlnQ

ΔG°reaction = ΔG°products - ΔG°reactants

ΔG°reaction = [( -23.4) +2(0)] - 3(-50.8)

ΔG°reaction = 129 kJ/mol

Q = PC3H8 * PH2^2/PCH4^3

Q = 0.013 * (2.3×10−2)^2/( 41)^3

Q = 6.877 * 10^-6/68921

Q= 9.9* 10^-11

Hence;

ΔG = 129 * 10^3 + [8.314 * 298 * (ln 9.9* 10^-11 )]

ΔG = 129 * 10^3 - 57073

ΔG = 72 kJ/mol

Learn more about free energy change: brainly.com/question/14143095

You might be interested in

Samantha AS

Anestetic [448]

Answer:

Fe(NO2)3

Explanation:

A poly atomic ion is an ion that contains more than one atom.

If we look at all the compounds that have been mentioned among the options; KBr and NaF are both ionic but do not contain any covalent poly atomic ion.

However, Fe(NO2)3 contains NO2^+ which is a poly atomic ion that contains the covalent bond.

8 0

3 years ago

Need help on number 14 please and thank u so much

mash [69]

Answer:

gravity

Explanation:

5 0

3 years ago

Define mass number of an atom.

Georgia [21]

The mass number of an atom is basically the total number of protons and neutrons.

6 0

2 years ago

3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?

Pavel [41]

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

$\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To balance it, we can simply add another sodium atom on the left. Hence:

$\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}$

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

$\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}$

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

$\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}$

And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

$\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Given the initial value and the above ratios, this yields:

$\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}$

Cancel like units:

$=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}$

Multiply. Hence:

$=0.3463...\text{ g H$_2$}$

Since we should have two significant values:

$=0.35\text{ g H$_2$}$

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

$\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}$

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

$\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}$

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

$\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}$

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

$\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}$

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: $\displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Using the given value and the above ratios, we acquire:

$\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}$

Cancel like units:

$\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}$