Which has greater mass?

Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac

valentinak56 [21]

Answer : The value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ = standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = $25^oC=273+25=298K$

Q = reaction quotient

First we have to calculate the $\Delta G_^o$ .

Formula used :

$\Delta G^o=-RT\times \ln K_p$

Now put all the given values in this formula, we get:

$\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})$

$\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole$

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

$CO(g)+2H_2(g)\rightarrow CH_3OH(g)$

The expression for reaction quotient will be :

$Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}$

$Q=8.1\times 10^{5}$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

Now put all the given values in this formula, we get:

$\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})$

$\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole$

Therefore, the value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

3 0

3 years ago

450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS

ExtremeBDS [4]

Answer:

1597.959 g

Explanation:

Given Data:

Amount of Cr₂(SO₄)₃ = 450 g

Amount of potassium phosphate K₃PO₄ = in Excess

grams of potassium sulfate K₂SO₄= ?

Solution

The Reaction will be

Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄

Information that we have from reaction

Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄

1 mol 2 mol 3 mol

we come to know from the above reaction that

1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄

We also know that

molar mass of Cr₂(SO₄)₃ = 147 g/mol

molar mass of K₃PO₄ = 212 g/mol

molar mass of K₂SO₄ = 174 g/mol

if we represent mole in grams then

Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄

1 mol (147 g/mol) 2 mol (212 g/mol) 3 mol (174g/mol)

So, Now we have the following details

Cr₂(SO₄)₃ + 2K₃PO₄ ----------> 3K₂SO₄ + 2CrPO₄

147 g 424 g 522 g

So,

we come to know that 147 g of Cr₂(SO₄)₃ combine with 424 g of 2K₃PO₄ produce 522 g of K₂SO₄

So now we calculate that how many grams of potassium sulfate will be produced

Apply unity formula

147 g of Cr₂(SO₄)₃ ≅ 522 g of K₂SO₄

450 g of Cr₂(SO₄)₃ ≅ ? g of K₂SO₄

by doing cross multiplication

g of K₂SO₄ =522 g x 450 g / 147 g

g of K₂SO₄ = 1597.959 g

So the write answer is 1597.959 g

***Note: By calculation it is obvious that the correct answer is 1597.959 g

8 0

3 years ago

Read the following chemical equations.

lora16 [44]

This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:

Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰

Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻

Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.

In such a way, the correct choice is C.

Learn more:

(Redox reactions) brainly.com/question/13978139

8 0

2 years ago

What is the pH of a solution with a concentration of 1.8 × 10-4 molar H3O+?

Andre45 [30]

PH is defined as the negative log of Hydrogen ion concentration. Mathematically we can write this as:

$pH=-log[H^{+}]=-log[H_{3}O]$

We are given the concentration of $H_{3}O$ . Using the value in formula, we get:

$pH=-log[1.8*10^{-4}]=3.745$

Therefore, the pH of the solution will be 3.745

8 0

3 years ago

A student found that the titration had taken 10.00 ml of 0.1002 m naoh to titration 0.132 g of aspirin, a monoprotic acid. calcu

Harman [31]

The balanced equation for the reaction between NaOH and aspirin is as follows;
NaOH + C₉H₈O₄ --> C₉H₇O₄Na + H₂O
stoichiometry of NaOH to C₉H₈O₄ is 1:1
The number of NaOH moles reacted - 0.1002 M / 1000 mL/L x 10.00 mL
Number of NaOH moles - 0.001002 mol
Therefore number of moles of aspirin - 0.001002 mol
Mass of aspirin reacted - 0.001002 mol x 180.2 g/mol = 0.18 g
However the mass of the aspirin sample is 0.132 g but 0.18 g of aspirin has reacted, therefore this question is not correct.

4 0

3 years ago

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