[huskywolf]

Hi , some help with this would be nice 

If the enthalpy of combustion of solid citric acid is – 1986 kJ mol-1, calculate the heat liberated when 10 g of solid citric acid undergoes total combustion at 298 K: (i) at constant pressure, (ii) at constant volume..

Now all I can think of to start is get the moles
n = m/M

the molar mass of anhydrous citric acid is 192.124 g/mol

n = 10g/ 192 g/mol

n = 0.052 mol

What do I do now to solve part (i) and (ii) please?

[Jesse2024]

You started correctly, now for constant pressure using these equations:

H = U +PV
U = q + w
dw = -PdV

You can come up with a relationship between enthalpy and heat, this would then mean a simple calculation.  I can help derive this if you need it.

For constant volume, I am not 100% sure on it yet.

Was there any other information given?

An isochoric and isothermal process (assuming the question meant constant temp since it only gives one) would mean that pressure is changing.

 :delta: H = q + nRTln(P₂/P₁)

But the question didn't give us any pressure values... I am a little confused.

[huskywolf]

You started correctly, now for constant pressure using these equations:

H = U +PV
U = q + w
dw = -PdV

You can come up with a relationship between enthalpy and heat, this would then mean a simple calculation.  I can help derive this if you need it.

For constant volume, I am not 100% sure on it yet.

Was there any other information given?

An isochoric and isothermal process (assuming the question meant constant temp since it only gives one) would mean that pressure is changing.

 :delta: H = q + nRTln(P₂/P₁)

But the question didn't give us any pressure values... I am a little confused.

Thanks for your help,
unfortunately thats all the information they give in the question?

[huskywolf]

Surely someone knows how to do this? 
Borek does 

[rabolisk]

First I would write a balanced chemical equation for the total combustion of citric acid. Since you are given data per mole (enthalpy of combustion), you have to scale it down because you only have 10 g of citric acid using molar mass. (You've already done this)

Now, there are three equations to use.

dU = dq + dw
dH = dU + d(pV)
dw = -pdV

These are differential equations which you have to integrate. For example, to get work, you have to integrate the third equation. For this problem, though, you won't have to worry about that. You can rewrite the second equation as
dH = dU + pdV + vdP
     = dq + dw + pdV + vdP
     = dq - pdV + pdV + vdP
     = dq + vdP

Use this to figure out the rest of part i. Part ii follows from this, except that pressure is not constant. Although the initial and final pressures are not given, you can solve this problem as long as you make one assumption about the behavior of the gases. The volume of the solid can be treated as negligible compared to that of the gases.