# Chemsitry Enthalpy Change Question Help

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When ethanoic acid reacts with sodium hydroxide, the enthalpy change, ∆H, is

–56.1 kJ mol–1

CH3COOH(aq) + NaOH(aq) ⟶ CH3COONa(aq) + H2O(l)

Calculate the temperature rise when 25 cm3 of 2.0 mol dm–3 aqueous ethanoic acid

react with 25 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide.

Assume that both solutions have the same initial temperature, have a density of

1.0 g cm–3 and a specific heat capacity of 4.18 J K–1 g–1

How would you do this? Obviously q=mcT

But would you work out moles, set out 1:1 ratio then use that as a mass of water? Help !

–56.1 kJ mol–1

CH3COOH(aq) + NaOH(aq) ⟶ CH3COONa(aq) + H2O(l)

Calculate the temperature rise when 25 cm3 of 2.0 mol dm–3 aqueous ethanoic acid

react with 25 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide.

Assume that both solutions have the same initial temperature, have a density of

1.0 g cm–3 and a specific heat capacity of 4.18 J K–1 g–1

How would you do this? Obviously q=mcT

But would you work out moles, set out 1:1 ratio then use that as a mass of water? Help !

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#2

The moles of both ethanoic acid and sodium hydroxide are = 0.05 mol. Due to 1:1 stoichiometry, this means that 0.05 mol of water are formed.

Since , this means that . Hence, Q = 56.1 * 0.05 = 2.805 kJ = 2805 J (the reason why I dropped the negative sign in the enthalpy change is because it simply shows that the enthalpy change is exothermic. You can't have negative energy, which would arise if I was to have kept the negative sign).

Now we have to use . Rearranging for would give . Let's make a checklist of the terms in the equation:

So, we have everything to calculate the temperature change. = 13.4 K.

Since , this means that . Hence, Q = 56.1 * 0.05 = 2.805 kJ = 2805 J (the reason why I dropped the negative sign in the enthalpy change is because it simply shows that the enthalpy change is exothermic. You can't have negative energy, which would arise if I was to have kept the negative sign).

Now we have to use . Rearranging for would give . Let's make a checklist of the terms in the equation:

- We calculated Q above; it's 2805 J.
- c is 4.18 J g
^{-1}K^{-1}. - m is 50 g. Since it's assumed that both solutions have a density of 1 g cm
^{-1}, and both solutions have volumes of 25 cm^{3}, this means they both have masses of 25 g. So, mixing these two solutions would give a combined volume of 50 cm^{3}, and so a combined mass of 50 g.

So, we have everything to calculate the temperature change. = 13.4 K.

Last edited by Kian Stevens; 1 year ago

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(Original post by

The moles of both ethanoic acid and sodium hydroxide are = 0.05 mol. Due to 1:1 stoichiometry, this means that 0.05 mol of water are formed.

Since , this means that . Hence, Q = 56.1 * 0.05 = 2.805 kJ = 2805 J (the reason why I dropped the negative sign in the enthalpy change is because it simply shows that the enthalpy change is exothermic. You can't have negative energy, which would arise if I was to have kept the negative sign).

Now we have to use . Rearranging for would give . Let's make a checklist of the terms in the equation:

So, we have everything to calculate the temperature change. = 13.4 K.

**Kian Stevens**)The moles of both ethanoic acid and sodium hydroxide are = 0.05 mol. Due to 1:1 stoichiometry, this means that 0.05 mol of water are formed.

Since , this means that . Hence, Q = 56.1 * 0.05 = 2.805 kJ = 2805 J (the reason why I dropped the negative sign in the enthalpy change is because it simply shows that the enthalpy change is exothermic. You can't have negative energy, which would arise if I was to have kept the negative sign).

Now we have to use . Rearranging for would give . Let's make a checklist of the terms in the equation:

- We calculated Q above; it's 2805 J.
- c is 4.18 J g
^{-1}K^{-1}. - m is 50 g. Since it's assumed that both solutions have a density of 1 g cm
^{-1}, and both solutions have volumes of 25 cm^{3}, this means they both have masses of 25 g. So, mixing these two solutions would give a combined volume of 50 cm^{3}, and so a combined mass of 50 g.

So, we have everything to calculate the temperature change. = 13.4 K.

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#4

(Original post by

Would you not calculate the mass of water produced in the reaction? I specifically remember being told that the mass is also the mass of water and not the mass of any solids/solutions?

**Principleturn**)Would you not calculate the mass of water produced in the reaction? I specifically remember being told that the mass is also the mass of water and not the mass of any solids/solutions?

^{3}of ethanoic acid and the 25 cm

^{3}of sodium hydroxide which are reacting and causing an enthalpy change.

What you were very much likely told was that the

**density**of a solution can be assumed to be the

**density**of water. The question wouldn't tell you to assume that the densities of both solutions are equal to that of water for nothing...

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(Original post by

No you wouldn't. The water is simply being formed, not undergoing any reaction; it's the 25 cm

What you were very much likely told was that the

**Kian Stevens**)No you wouldn't. The water is simply being formed, not undergoing any reaction; it's the 25 cm

^{3}of ethanoic acid and the 25 cm^{3}of sodium hydroxide which are reacting and causing an enthalpy change.What you were very much likely told was that the

**density**of a solution can be assumed to be the**density**of water. The question wouldn't tell you to assume that the densities of both solutions are equal to that of water for nothing...
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#6

i don't understand why you do the enthalpy change multiplied by 0.05, why are we altering the value they have given us? also i dont get why at the end the masses combine to get the answer in cm^3 and why we haven't divided by 100?this is what i would do , i'm assuming its wrong but i don't know why.5.6.1/0.1 x 4.18 = 134.21...any help is appreciated

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