Sunday

March 29, 2015

March 29, 2015

Posted by **Mandy** on Sunday, November 29, 2009 at 6:37pm.

You may only use the following information:

N2 (g) + O2 (g) into 2 NO (g); Delta Hf = 180.6 kJ

N2 (g) + 3 H2 (g) into 2 NH3 (g); Delta Hf = -91.8 kJ

2 H2 (g) + O2 (g) into 2 H2O (g); Delta Hf = -483.7 kJ

My answer was -910.6 KJ but the real answer was -918.3 KJ. I need to know what I am doing wrong.

- Chemistry II -
**DrBob222**, Sunday, November 29, 2009 at 8:28pmI didn't get either answer.

Look at your equation to make sure it adds to the desired equation in the problem. Here is what I did.

equation 1 x 2

equation 2 reversed and x 2.

equation 3 x 3.

Add it. First look at the equation.

(2N2) + 2O2 + 4NH3 + (6H2) +3O2 ==>

(2N2) + + (6H2) + 4NO + 6H2O

The 2N2 and 6H2 cancel since they are on opposite sides of the equation and we are left with

4NH3 + 5O2 ==>4NO + 6H2O which is exactly what we want.

Then (180.6*2)+(91.8*2)+(-483.7*3) = -906.3 kJ.

Check my work.

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