Question

Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl4(g)] = −96.0 kJ/mol ΔH o f [CCl4(l)] = −139 kJ/mol ΔH o f [HCl(g)] = −92.31 kJ/mol ΔH o f [HCl(aq)] = −167.46 kJ/mol ΔH o f [Cl(g)] = 121.0 kJ/mol

Answer 1

Answer:  ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]$

$\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]$

$\Delta H=-277.4kJ$

Therefore, the enthalpy change for this reaction is, -277.4 kJ