Chemistry, 06.05.2020 02:19 becddjr

When 22.7 mL of 0.500 M H2SO4 is added to 22.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)___kJ/mol H2O

Answers: 1

Show answers

Answers

Answer from: bigdramm409

Answer : The enthalpy of neutralization is, -55.8 KJ/mole

Explanation :

First we have to calculate the moles of H_2SO_4 and KOH.

$\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.500mole/L\times 0.0227L=0.01135mole$

$\text{Moles of KOH}=\text{Concentration of KOH}\times \text{Volume of solution}=1.00mole/L\times 0.0227L=0.0227mole$

Now we have to calculate the limiting reagent.

The balanced chemical reaction will be,

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of H_2SO_4 react to give 2 moles of H_2O

So, 0.01135 mole of H_2SO_4 react to give $2\times 0.01135=0.0227$ moles of H_2O

and,

As, 2 moles of KOH react to give 2 moles of H_2O

As, 0.0227 moles of KOH react to give 0.0227 moles of H_2O

From this we conclude that they can form the same amount of product. This means that both will consume completely in the reaction.

Thus, the number of neutralized moles = 0.0227 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 22.7ml+22.7ml=45.4ml

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 45.4ml=45.4g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

= specific heat of water = 4.184J/g^oC

m = mass of water = 45.4 g

$T_{final}$ = final temperature of water = 30.17^oC

$T_{initial}$ = initial temperature of water = 23.50^oC

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

$q=45.4g\times 4.184J/g^oC\times (30.17-23.50)^oC$

q=1266.99J=1.267kJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -1.267 KJ

n = number of moles used in neutralization = 0.0227 mole

$\Delta H=\frac{-1.267KJ}{0.0227mole}=-55.8KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, -55.8 KJ/mole

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

The answer is probably c. xylem

Another question on Chemistry

Chemistry, 23.06.2019 04:00

Why must humans find substitutes for many minerals found on earth? (a) form at an extremely slow rate (b) controlled by other countries (c) too deep in the earth to collect

Answers: 1

Answer

Chemistry, 23.06.2019 10:20

El amoniaco y el fluor reaccionan para formar tetrafluoruro de dinitrogeno y fluoruro de hidrogeno. segun la reaccion: nh3 + f2 ⇒ n2f4 + hf si reaccionan 5 gramos de amoniaco y 20 gramos de fuor, ¿cuantos gramos de fluoruro de hidrogeno se producen?

Answers: 2

Answer

Chemistry, 23.06.2019 10:30

Using the periodic table, complete the following. element: hydrogen symbol: h₂ molecular weight: g mass of one mole: g/mol

Answers: 2

Answer

Chemistry, 23.06.2019 23:00

In the chemical formula 3h2 identify the subscript identify the coefficient how many atoms of hydrogen are present

Answers: 3

Answer

You know the right answer?

When 22.7 mL of 0.500 M H2SO4 is added to 22.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50...