# Calculate the enthalpy of the reaction: 2NO(g) + O2(g) –> 2NO2(g)?

0

Calculate the enthalpy of the reaction: 2NO(g) + O2(g) –> 2NO2(g)

given the following reactions and enthalpies of formation:

1/2 N2(g) + O2(g) –> NO2(g), delta H *A = 33.2 kJ

1/2 N2(g) + 1/2O2(g) –> NO(g), delta H *B= 90.2 kJ

Calculate the enthalpy of the reaction: 4B(s)+3O2(g) –> 2B2O3(s)

given the following pertinent information:

B2O3(s) + 3H2O(g) –> 3O2(g) + B2H6 (g) delta H *A = +2035 kJ

2B(s) + 3H2(g) –> B2H6(g) delta H *B = +36 kJ

H2(g) + 1/2O2(g) –> H2O(l) delta H *C = -285 kJ

H2O(l) –> H2O(g) delta H *D = +44 kJ

Can someone show me step by step? I would be very grateful because I do not understand Hess’s law very well.

• The Hess law is used to discover the reaction’s entalphy and doing it, you must cancel the same substances, if they are in differents sides, to make the given reaction. Remember that if you multiply or divide the reaction, the same happens with the entalphy and when you invert the reaction, the entalphy’s sign must be also inverted. The first one is easy, but the second one took me some work.

1)

The reaction you have is: 2 NO + O2 -> 2 NO2

It means that 2 NO and O2 must be on the left side and 2 NO2, on the right one, so you will look for the entalphies that were given and invert and/or multiply to copy this order, look:

1/2 N2(g) + O2(g) –> NO2(g), deltaH = 33.2 kJ (should be multiplied by 2)

N2 + 2 O2 -> 2 NO2………deltaH = +66,4 kJ

1/2 N2(g) + 1/2O2(g) –> NO(g), delta H = 90.2 kJ (should be multiplied by 2 and inverted)

2 NO -> N2 + O2…………..deltaH = -180,4 kJ

Now you have:

N2 + 2 O2 -> 2 NO2………deltaH = +66,4 kJ

2 NO -> N2 + O2…………..deltaH = -180,4 kJ

canceling both N2 and 2 O2 with O2, you will find the given reaction:

2 NO + O2 -> 2 NO2……..deltaH = 66,4 – 180,4 = -114,0 kJ

2)

In this one, you must know that you cannot cancel, for example, H2O(l) with H2O(g). The substances must be in the same physical state.

The reaction you have is: 4 B(s) + 3 O2(g) –> 2 B2O3(s)

It means that you have 4B(s) and 3 O2(g) on the left side and 2 B2O3(s) on the right side.

I will start for the second given entalphy reaction.

2 B(s) + 3 H2(g) –> B2H6(g), deltaH = +36 kJ (you need 4 B, so you should multiply this by 2)

4 B(s) + 6 H2(g) -> 2 B2H6(g)…….deltaH = +72 kJ

B2O3(s) + 3 H2O(g) –> 3 O2(g) + B2H6(g), deltaH = +2035 kJ

at first look you need just to invert because of the 3 O2, but you have another equation that involves oxygen, so you will see that here you need to invert AND multiply by 2 (you can also look at the B2H6, where in the reaction above you have 2, so now you need 2 to cancel, as they are on different sides)

6 O2(g) + 2 B2H6(g) -> 2 B2O3(s) + 6 H2O(g)..deltaH = -4070 kJ

H2(g) + 1/2 O2(g) –> H2O(l), deltaH = -285 kJ

Since you have 6 O2 and you must have only 3 O2, you should now multiply by 6 and invert this reaction:

6 H2O(l) -> 6 H2(g) + 3 O2(g)……..deltaH = +1710 kJ

H2O(l) –> H2O(g), deltaH = +44 kJ

To cancel both H2O(l), you should multiply it by 6 and also invert

6 H2O(g) -> 6 H2O(l) ……deltaH = -264 kJ

Now you have:

6 O2(g) + 2 B2H6(g) -> 2 B2O3(s) + 6 H2O(g)..deltaH = -4070 kJ

4 B(s) + 6 H2(g) -> 2 B2H6(g)……………………….deltaH = +72 kJ

6 H2O(l) -> 6 H2(g) + 3 O2(g)………………………deltaH = +1710 kJ

6 H2O(g) -> 6 H2O(l)………………………………….deltaH = -264 kJ

doing the same way as we did before, you will find the same given reaction:

4 B(s) + 3 O2(g) -> 2 B2O3(s)

deltaH = 72 + 1710 – 4070 – 264 = -2552 kJ

• Use the H values for equations 1 and 2 to determine the value of H for equation 3.

(1) 2NO N2 + O2 H = −180. kJ

(2) 2NO + O2 2NO2 H = −112 kJ

(3) N2 + 2O2 2NO2 H = ?

RE:

Calculate the enthalpy of the reaction: 2NO(g) + O2(g) –> 2NO2(g)?

Calculate the enthalpy of the reaction: 2NO(g) + O2(g) –> 2NO2(g)

given the following reactions and enthalpies of formation:

1/2 N2(g) + O2(g) –> NO2(g), delta H *A = 33.2 kJ

1/2 N2(g) + 1/2O2(g) –> NO(g), delta H *B= 90.2 kJ

Calculate the enthalpy of the reaction: 4B(s)+3O2(g) –>… 