# Enthalpy of reaction between ammonia and oxygen

I have to ask this here because I asked literally everybody I know and I still disagree with them.

Using the information in the table below, calculate the heat of released from the following reaction in $$\pu{kJ}$$ per $$\pu{g}$$ of $$\ce{NO}.$$ (Mw. of $$\ce{NO}$$ is $$\pu{30.01 g/mol})$$

$$\ce{4 NH3(g) + 5 O2(g) -> 6 H2O(g) + 4 NO(g)}$$

$$\begin{array}{lrrr} \hline & \ce{NH3} & \ce{H2O} & \ce{NO} \\ \hline H_\mathrm{f}^\circ~(\pu{kJ/mol}) & -46.1 & -241.8 & +90.3 \\ \hline \end{array}$$

a) $$-13.29$$
b) $$-8.372$$
c) $$-7.541$$
d) $$-30.16$$
e) $$-5.678$$

I'm convinced the answer is c.

The enthalpy of this reaction for 4 moles of NO is -905.2, to find the enthalpy for 1 mole of NO we have to divide by 4 to get to -226.3.

Then, to find it per gram of NO we have to divide by 30.01, this gives us 7.541.

Am I right ? If not, what did I do that was wrong? Where's my mistake?

Please don't put this as HW help, I asked multiple websites and went through my working 5-6 times but I still couldn't point out where I went wrong.

• It seems to me that c is correct. – Chet Miller Jan 2 at 16:40
• sorry to interrupt, do you 3 people agree its C ? – Jaja bae Jan 2 at 18:32
• Yes, I agree with C – Zenix Jan 2 at 18:38
• I also agree with C. – Maurice Jan 2 at 19:35
• How did you know three people agree before the three people commented? Or were comments deleted? – Buraian Oct 9 at 22:22

The heat of reaction of a mixture is defined as the extent of reaction, $$\xi$$ times the enthaply of reaction, $$\Delta H_{rxn}$$,

$$Q_{rxn} =\xi\Delta H_{rxn} = \sum^n_{i=1}\nu_iH_{f,i}^{o} =\sum^{n,products}_{i=1}\|\nu_i\|H_{f,i}^{o} - \sum^{m,reactants}_{i=1}\|\nu_i\|H_{f,i}^{o}$$

For a reaction system,

$$\ce{4NH_3(g) + 5O_2(g) -> 6H_2O(g) + 4NO(g)}$$

In the case of the example you've provided, you are assuming this reaction is taking place at standard conditions. If we assume for a stiociometic amount of reactants, then the extend of reaction, $$\xi$$, should be equal to 1. Therefore,

$$Q_{rxn} = 1*\Delta H_{rxn} = 6\Delta H^o_{f,H_2O} + 4\Delta H^o_{f,NO} - 5\Delta H^o_{f,O_2} - 4\Delta H^o_{f,NH_3}$$

Since, $$\Delta H^o_{f,O_2} = 0 \pu{kJ//mol}$$, the equation reduces to,

$$\Delta H_{rxn} = 6\pu{mol}(-241.8 \pu{kJ//mol}) + 4\pu{mol}(90.3 \pu{kJ//mol}) - 4\pu{mol}(-46.1 \pu{kJ//mol})$$

$$\Delta H_{rxn}= -905.2\pu{kJ}$$

Now, per gram of $$\ce{NO}$$, assuming the basis of 4 mol $$\ce{NO}$$ from the reaction equation,

$$\frac{Q_{rxn}}{m_{NO}} = \frac{\Delta H_{rxn}}{m_{NO}} = \frac{-905.2\pu{kJ}}{4_{mol, NO}(30.01\pu{g//mol})}=-7.541\pu{kJ//g_{,NO}}$$

• You've used the wrong units and it confused me. $$\Delta H_{rxn}= -905.2\frac{kJ}{mol}$$ The enthaply of reaction is just regular enthalpy change with units $\pu{kJ}$. Your unit had per mole in it. I ask you, per mole of what substance? I'm 100 percent sure that the OP's friends did the exact same mistake. It is the enthaply of formation, combustion etc. which have units define for 1 mole the product. Enthalpy of reaction has units $\pu{kJ}$ – Eyy boss Oct 10 at 17:50
• Also, when you multiplied $\Delta H_f^\circ$ (which has unit $\pu{kJ/mol}$) by the respective number of moles of the substance the unit $\pu{mol}$ got cancelled. Example: $$6\Delta H^o_{f,H_2O} = 6(-241.8 \frac{kJ}{mol})$$ you should have written the unit $\pu{mol}$ with the number $6$. Which would've lead to $$\pu{mol}\times\pu{\frac{kJ}{mol}} = \pu{kJ}$$ – Eyy boss Oct 10 at 17:59