Question

Calculate ΔHrxn for the following reaction: CH4(g)+2Cl2(g)→CH2Cl2(g)+2HCl(g) Use the following reactions and given ΔH values. CH4(g)+Cl2(g)→CH3Cl(g)+HCl(g), ΔH=−99.60 kJ CH3Cl(g)+Cl2(g)→CH2Cl2(g)+HCl(g), ΔH=−105.8 kJ Express your answer to four significant figures.

Answer 1

Answer: -205 kJ

Step-by-step explanation:

Since this represents two consecutive processes to yield a particular product, we can see that nothing needs to be done to either equation. They can simply be added together as they are, and things will cancel to yield the desired product.

−99.6 kJ–105.8 kJ=−205 kJ

Answer 2

Answer: The
`\Delta H_(rxn)` for the reaction is -205.4 kJ.

Step-by-step explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of methane and chlorine gas follows:

`CH_4(g)+2Cl_2(g)\rightarrow CH_2Cl_2(g)+2HCl(g)`
`\Delta H_(rxn)=?`

The intermediate balanced chemical reaction are:

(1)
`CH_4(g)+Cl_2(g)\rightarrow CH_3Cl(g)+HCl(g)`
`\Delta H_1=-99.60kJ`

(2)
`CH_3Cl(g)+Cl_2(g)\rightarrow CH_2Cl_2(g)+HCl(g)`
`\Delta H_2=-105.8kJ`

The expression for enthalpy of the reaction follows:

`\Delta H_(rxn)=[1* \Delta H_1]+[1* \Delta H_2]`

Putting values in above equation, we get:

`\Delta H_(rxn)=[(1* (-99.60))+(1* (-105.8))]=-205.4kJ`

Hence, the
`\Delta H_(rxn)` for the reaction is -205.4 kJ.