Question

Isobutanol (C4H10O) is anorganic alcohol that is used mainly as a solvent in organic reactions,but can also be used to produce diesel and jet fuel. The first reaction in the manufacturing of diesel isthe production of 2-methylpropene (C4H8): C4H10O(l) ➝C4H8(g) + H2O(g). Use the chemical reactions below to calculate the enthalpy change (ΔHrxn) for the production of 2-methylpropene at 298K.4C(s)+ 5H2(g) + ½O2(g) ®C4H10O(l)DHo= -335 kJ/mol4C(s)+ 4H2(g) ®C4H8(g)DHo= -17 kJ/mol2H2(s)+ O2(g) ®2H2O(g)DHo= -484kJ/mol

Answer 1

answer =76KJ/mol

explanation :

​​​​​​First we shall write down the Equation which are given to solve the problem.

Eq.1. C4H10O(l)------------> C4H8(g) + H2O(g) ,∆H = ?

The given chemical reaction are ,

EQ.2. 4C(s) + 5H2(g) + 1/2 O2(g) ------> C4H10O(l) , ∆H = -335 KJ/mol.

EQ.3. 4C(s) + 4 H2(g) ----------> C4H8(g) , ∆H = -17 KJ/mol.

EQ. 4 2H2(g) + O2(g) --------> 2H2O(g) , ∆H = -484 KJ/mol.

Now, we shall arrange the above equation in such a way that on adding ,and applying Hess law of summation heat enthalpy we get the overall Equation for the formation of 2-methyl propene.

Reverse the Equation 2 and re write also multiply by 2. ,the sign of enthalpy will change from negative to positive and multiplication by 2 also takes place,

EQ.5. 2C4H10O(l) -----------> 8C(s) + 10H2(g) + O2 (g) , ∆H = 2* 335 = 670 KJ/mol.

Multiplying the Equation 3 by 2 and re write,

EQ.6, 8C(s) + 8H2(g) ----------> 2 C4H8(g) , ∆H= 2*(-17) = -34 KJ/mol.

Rewrite the equation 4 ,as same naming eq. 7

EQ. 7,. 2H2(g) + O2(g) --------> 2H2O(g) , ∆H = -484 KJ/mol

Now, Adding Equation 5,6 and 7 , as well as enthalpies.

we find,

2C4H10O(l) -----------> 2C4H8(g) + 2H2O(g) , ∆H =670-34-484= 152 KJ/mol.

Deviding above Equation by 2, and then writing -

C4H10O(l) ---------> C4H8(g) + 2H2O(g) , ∆H = 76 KJ/mol.

​​​​​​Therefore the enthalpy of formation of 2-methyl propene will be 76KJ/mol.

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