All categories

3 years ago

An irreversible reaction is one that:

Chemistry

1 answer:

bazaltina [42]3 years ago

5 0

Answer:

(d) has an extremely large equilibrium constant.

Explanation:

Hello,

Considering a generic chemical reaction:

$aA+bB cC+dD$

The equilibrium constant is defined as:

$K=\frac{[C]^c_{eq}[D]^d_{eq}}{[A]^a_{eq}[B]^b_{eq}} \\$

Now, an irreversible chemical reaction is a reaction in which the reagents are converted into products with no chance of coming back, so, considering the previous chemical reaction, the concentration of both A and B tends to be zero, so an extremely large equilibrium constant is gotten.

Best regards.

You might be interested in

The reaction between bromine gas and fluorine gas to create bromine monofluoride gas has reached equilibrium. what is the effect

WITCHER [35]

More bromine gas will be produced.

5 0

3 years ago

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol$

To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0

3 years ago

Observation And Assessment

Sloan [31]

I'm confused here??????

5 0

3 years ago

Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate

Anvisha [2.4K]

Answer:

$-3272$ kJ/mol

Explanation:

Given and known facts

Mass of Benzene $= 0.187$ grams

Mass of water $= 250$ grams

Standard heat capacity of water $= 4.18$ J/g∙°C

Change in temperature ΔT $= 7.48$ °C

Heat

$=250 * 4.18 * 7.48\\=7816.6 \\=7.82$

Heat released by benzine is - 7.82 kJ

Now, we know that

$0.187$ grams of benzene release $= -7.82$ kJ heat

So, $1$ g benzine releases

$\frac{ -7.82 }{0.187}\\= -41.8$

kJ/g

$0.187 * \frac{1}{78.108}=0.00239$ mol C6H6

Heat released

$= \frac{-7.82}{ 0.00239}$

$=-3272$ kJ/mol

4 0

3 years ago

0.023 M solution of perchloric acid

kogti [31]

The original question is to find the pH and the pOH of 0.023 M of perchloric acid.

Answer:
pH = 1.638
pOH = 12.362

Explanation:
1- getting the pH:
pH can be calculated using the following rule:
pH = -log[H+]
Since the given solution is an acid, this means that [H+] is the same as the concentration of the solution.
This means that:
[H+] = 0.023
Substitute in the above equation to get the pH as follows:
pH = -log[0.023]
pH = 1.638

2- getting the pOH:
We know that:
pH + pOH = 14
We have calculated that pH = 1.638.
Substitute in the above equation to get the pOH as follows:
pOH + 1.638 = 14
pOH = 14 - 1.638
pOH = 12.362

Hope this helps :)

5 0

3 years ago