Given the standard enthalpy changes for the following two reactions:
(1) n2(g) + 2o2(g)2no2( δh° = 66.4 kj
(2) 2n2o(g)2n2(g) + o2(δh° = -164.2 kj
what is the standard enthalpy change for the reaction:
(3) 2n2o(g) + 3o2(g)4no2(δh° =

The answer is -297 kJ

Explanation:

The enthalpy change is state function and can be calculated from Hess' s Law

The two reactions can be combined to obtain the third equation to calculate its enthalpy change as the enthalpy change depends upon only the initial and final state irrespective of path taken.

(1) ...... ΔH° = 66.4 kJ

(2) ......ΔH° = -164.2 kJ

The required equation is:

(3) ......ΔH° = _________?

We will multiply the equation (1) with "2" and then will subtract it from equation "1"

(2) ......ΔH° = -164.2 kJ

-2[X(1) ...... ΔH° = 66.4 kJ ]

......................................................................................................................................

ΔH° = ΔH°(eq 2) - 2ΔH°(eq 1) = -164.2 -(2X66.4) = -297 kJ

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"