N2(g) + 2 O2(g) 2 NO2(g) ΔH = +66.4 kJ

What would the enthalpy change for the following reaction be?

4 NO2(g) 2 N2(g) + 4 O2(g) ΔH = ???

a. +66.4 kJ
b. -66.4 kJ
c. +132.8 kJ
d. -132.8 kJ
e. -33.4 kJ

The answer to your question is the letter d. -132.8 kJ

Explanation:

To solve this problem, just observe that the ΔH is originally equal to +66.4 kJ.

If we change the order of the reaction, ΔH changes its sign but the value is the same.

Finally, if we duplicate the number of moles, also the ΔH increases twice.

For this problem

                           ΔH = 2(-66.4 kJ)

                          ΔH = - 132.8 kJ

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally