Question

Example 41. Calculate the standard heat of formation of
carbon disulphide (1). Given that the standard heats of combus-
tion of carbon (s), sulphur (s) and carbon disulphide (1) are
-393.3, 293.7 and -1108.76 kJ mol-respectively.​

Answer 1

Answer:

the answer is 128.06 KJ mol-1

Explanation:

We have to find :

C(s) + 2S(s) -----> CS2(l) ΔHf = ?

Given:

1. C(s) + O2(g) --> CO2(g) ΔH = -393.3 kJ

2. SO2(g) ----> S(s) + O2 ΔH = 293.7 kJ

3. CS2(l) + 3O2(g) ----> CO2(g) + 2SO2(g) ΔH = -1108.76 kJ

Now,

Keep 1.,

double 2. and reverse it,

and reverse 3.

and add them.

ONE OF THE MOST IMPORTANT POINT IS:-

• If you double an equation, u have to double the ΔH value and
• If you reverse an equation u have to change the sign of ΔH

We have now 3 different equn.s

4. C(s) + O2(g) --> CO2(g) ΔH = -393.3 kJ

5. 2S(s) + 2O2(g) ---> 2SO2(g) ΔH = 2 x -293.7 kJ = -587.4 kJ

6. CO2(g) + 2SO2(g) --> CS2(l) + 3O2(g) ΔH = +1108.76 kJ

Adding 4,5 and 6 gives:

C(s) + O2(g) + 2S(s) + 2O2(g) + CO2(g) + 2SO2(g) --> CO2(g) + 2SO2(g) + CS2(l) + 3O2(g)

C(s) + 2S(s)---->CS2

Therefore,ΔH = -393.3 + (-587.4) + 1108.76 kJ

By solving we get

ΔH = 128.06 kJ.