Aconstant external pressure of 43.6 bar, a mole of x reacts from the reaction x + 4y = 2z. before the reaction the volume of the gaseous mixture was 5.00 l .after the reaction the volume 2.00 l. calculate the value if the total energy, δu in kj/mol. δh °=75.0kj/mol.

ΔU = 88.8 KJ/mol

Explanation:

to Calculate  ΔU we are going to use this formula:

ΔU = q + w

when

ΔU is change in internal energy

and q is the heat added to the system

and W is the work done by the system

at constant pressure :

we can get w from this formula:

w= -PΔV

and when p is the pressure = 43.6 bar we need to convert bar to N/m^2

1 N/m^2 = 1 * 10^-5 bar

So 43.6 bar = 4360000 N/m^2

now ΔV is the change in volume = Vfinal - Vinitial

ΔV = 2L - 5  L = - 3 L

but we need to convert this unit to m^3

1 litre = 0.001 cubic meters

so ΔV = 3 L * 0.001 = - 0.003 m^3

by substitution in work formula:

w = - 4360000 N/m^2 *- 0.003 m^3

   = 13080 N m

N m unit = Joule and to convert it to kJ

1 Joule = 0.001 KJ

So W = 13080 * 0.001

         = 13.08 KJ

now we need to know q , at constant pressure q = ΔH ° = 75 KJ/mol

So by substitution in ΔU formula:

ΔU  = 75 KJ/mol +13.08 KJ

       = 88.8 KJ/mol  

h let me think

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