Question

chemist measures the enthalpy change ΔΗ during the following reaction: COC12(g) + 4NH3(g) CO (NH 2),(s) + 2 NH4Cl(s) ΔΗ.-559. kl Use this information to complete the table below. Round each of your answers to the nearest kJ/mol. reaction 559. kJ kJ NH (NH)+ NH,CO) kJ

Answer 1

Hi,

The given reaction is:

$COCl_2(g) + 4NH_3(g) \rightarrow CO(NH_2)_2+2NH_4Cl(s)$ with $\Delta H = -559 kJ$

Now, first entry is just the reverse of the above reaction which results in just a positive value of Delta H .

The second entry in the table is just the first entry multipied by 3. Since the reported value of Delta H is not molar hence Delta H will change exactly with this multiplication factor. Hence,

For entry 2, $\Delta H = 3*559 kJ$ = 1677 kJ

Now similarly, entry 3 can be obtained by multiplying entry 1 by 1/2, hence, its Delta H should also be multiplied by the same facotr.

So, for entry 3,  $\Delta H = \frac{1}{2}*559 kJ$  

= 279.5 kJ