Question

Calculate ΔrH for the following reaction:

CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g)

Use the following reactions and given ΔrH's.

C(s)+2H2(g)→CH4(g)ΔrH=−74.6kJmol−1C(s)+2Cl2(g)→CCl4(g)ΔrH=−95.7kJmol−1H2(g)+Cl2(g)→2HCl(g)ΔrH=−92.3kJmol−1

Answer 1

Ans:

= Enthalpy of reaction = Heat change (absorbed or released) of the reaction at constant pressure

can be calculated by using following formula.

Hess's Law of constant heat summation : Enthalpy change of reaction is constant , irrespective of number of steps or methods of reaction.

e.g. Following two reactions have same irrespective of their steps.

i) A --> B --> C

ii) A --> C Here,

Reaction whose Enthalpy change is to be calculated i.e. TARGET REACTION,

CH_4(g)+4Cl_2(g)→CCl_4(g)+4HCl_(g)    = ?

Thermochemical equation is Balanced chemical equation along with its value.

Now, According to Hess's law we have to arrange the thermochemical equations and sum up them to give our TARGET REACTION.

C_(s)+2H_2(g)→CH_4(g) ΔrH = −74.6 kJmol−1 ---------------------------(1)   

C_(s)+2Cl_2(g)→CCl_4(g) ΔrH = −95.7 kJmol−1 ---------------------------(2)

H_2(g)+Cl_2(g)→2HCl_(g) ΔrH = −92.3 kJmol−1 ----------------------------(3)

Reaction whose Enthalpy change is to be calculated i.e. TARGET REACTION,

CH_4(g)+4Cl_2(g)→CCl_4(g)+4HCl_(g)    = ?

To achieve target equation we have to do some operations on eq. 1, 2 and 3 and do summation of them.

1) Reverse the eq. 1 (reversing equation changes the sign of ΔrH)

CH_4(g)   →  C_(s)+2H_2(g) ΔrH = +74.6 kJmol⁻¹ ---------------------------(1)   

2) keep eq. 2 as it is

C_(s)   + 2Cl_2(g) → CCl_4(g) ΔrH = −95.7 kJmol⁻¹ ---------------------------(2)

3) Multiply (eq. 3) by 2 and finally sum up these to get final equation.

( H_2(g) + Cl_2(g) → 2HCl_(g) ΔrH = −92.3 kJmol^₋₁ ) x 3 ----------------------(3)

And finally sum up these thermochemical equations to get final Thermochemical equation.

CH_4(g)   → C_(s)+2H_2(g) ΔrH₁= +74.6 kJmol⁻¹ ---------------------------(1)   

C_(s)   + 2Cl_2(g) → CCl_4(g) ΔrH₂= −95.7 kJmol⁻¹ ---------------------------(2)

2H_2(g) + 2Cl_2(g) → 4HCl_(g) ΔrH₃= −184.6 kJmol^₋₁ ----------------------------(3)

  

CH_4(g)  + 4Cl_2(g)   → CCl_4(g) + 4HCl_(g) ΔrH = −205.7 kJmol^₋₁ ------------------(Final eq.)

[Note: During Summation cancel out common chemicals from LHS and RHS of the equation and also sum up the ΔrH ]