Question

a constant 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.

Answer 1

To calculate the total change in internal energy (ΔU), use the equation ΔU = ΔH - PΔV. Plugging in the given values, the value of ΔU is approximately 206.2 kJ/mol.

Step-by-step explanation:

To calculate the total change in internal energy (ΔU), we can use the equation ΔU = ΔH - PΔV, where ΔH is the change in enthalpy, P is the external pressure, and ΔV is the change in volume. In this case, ΔH is given as 75.0 kJ/mol, P is 43.6 bar (convert to atm), and ΔV is the difference between initial and final volumes (convert to L). Plugging in the values, we can calculate ΔU. Remember to convert the final answer to kJ/mol.

ΔU = ΔH - PΔV
ΔU = 75.0 kJ/mol - (43.6 atm * (2.00 L - 5.00 L))
ΔU ≈ 75.0 kJ/mol - (-131.2 kJ)

ΔU ≈ 206.2 kJ/mol

Answer 2

ΔU = 88.8 KJ/mol

Step-by-step 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