Chemistry, 15.04.2020 00:47 camdynmmm

In a coffee‑cup calorimeter, 65.0 mL of 0.890 M H 2 SO 4 was added to 65.0 mL of 0.260 M NaOH . The reaction caused the temperature of the solution to rise from 23.78 ∘ C to 25.55 ∘ C. If the solution has the same density as water (1.00 g/mL) and specific heat as water (4.184 J/g‑K), what is Δ H for this reaction (per mole of H 2 O produced)? Assume that the total volume is the sum of the individual volumes.

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Answer from: zaire8

Answer : The enthalpy of neutralization is, 56.96 kJ/mole

Explanation :

First we have to calculate the moles of H₂SO₄ and NaOH.

$\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.890mole/L\times 0.065L=0.0578mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.260mole/L\times 0.065L=0.0169mole$

The balanced chemical reaction will be,

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

From the balanced reaction we conclude that,

As, 2 mole of NaOH neutralizes by 1 mole of H₂SO₄

So, 0.0169 mole of NaOH neutralizes by 0.00845 mole of H₂SO₄

That means, NaOH is a limiting reagent and H₂SO₄ is an excess reagent.

Now we have to calculate the moles of H₂O.

As, 2 mole of NaOH react to give 2 mole of H₂O

So, 0.0169 mole of NaOH react to give 0.0169 mole of H₂O

Now we have to calculate the mass of water.

As we know that the density of water is 1.00 g/ml. So, the mass of water will be:

The volume of water = 65.0ml+65.0ml=130ml

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/ml\times 130ml=130g$

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 = 130 g

$T_{final}$ = final temperature of water = 23.78^oC

$T_{initial}$ = initial temperature of metal = 25.55^oC

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

$q=130g\times 4.184J/g^oC\times (25.55-23.78)^oC$

q=962.7J

Thus, the heat released during the neutralization = -962.7 J

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -962.7 J

n = number of moles used in neutralization = 0.0169 mole

$\Delta H=\frac{-962.7J}{0.0169mole}=-56964.49J/mole=-56.96kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.96 kJ/mole

Answer from: Quest

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .

Answer from: Quest

answer: any goods being imported from another country may be subject to issued by the federal government are tariffs

explanation:

tariffs are a tax on anything that are imported from abroad. government use them to protect and boost their local industries and the thought is that they encourage consumers to buy local instead of importing and they will hope will protect jobs and economy in the country that implements them

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In a coffee‑cup calorimeter, 65.0 mL of 0.890 M H 2 SO 4 was added to 65.0 mL of 0.260 M NaOH . The...