Question

calculate the standard enthalpy of formation of so3 at 298k using the following reactions and enthalpies S2+8o2=8so2 dH=-2775 KJ mol-1 2So2+O2=2So3 dH=-198 KJ mol-1​.

ANSWER THIS QUESTION WITH FULL EXPLANATION ✌️​

Answer 1

Given:

T = 298 k

4S2 + 8O2 ---> 8SO2 ∆H = -2775KJ/mol ( equation 1)

((note: Here 4 moles of S2 are required in the question it is not mentioned but for reaction to take place it is mandatory))

2SO2 + O2 ---> 2SO3∆H = -198 KJ/mol (equation 2)

To Find:

The heat of formation of SO3

Solution:

The reaction of formation of SO3 is as follows:

$\frac{1}{2} s2 \: + \frac{3}{2} o2 = so3$

(equation 3)

From the above equation 3, we need S2 and O2 as reactants, which we get in equation 1.

To equalize the number of moles we divide equation 1 by 8,

we get;

½S2 + O2 ---> SO2 (3)

Also, the heat energy for equation 1 will be divided

$\frac{ - 2775}{8} = - 346.876kj$

Now, dividing the equation 2 by 2 we get,

SO2 + ½O2 ---> SO3 (4)

Therefore, the heat of the reaction also gets divided by 2.

we get,

$\frac{ - 198}{2} = - 99kj$

Now adding (3) and (4)

we get,

∆H = - 346.876kj

+ ∆H = - 99.00 KJ

= -445.875 KJ/mol

Therefore, the heat of formation of SO3 is ∆H = -445.875kj/mol.