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

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.