Question

Data: A H f values: CH 4( g), -74.8 kJ; CO 21 g), -393.5 kJ; H 20( 1), -285.8 kJ. Using the A H f data above, calculate A H xn for the reaction below. Reaction: CH 4( 9) + 20 2( 9) => CO 2(g) + 2H 2011) Selected Answer: d. -890.3 kJ Correct Answer: d. -890.3 kJ

Answer 1

`\Delta H_(rxn)` for the given reaction is -890.3 kJ

Step-by-step explanation:

`\Delta H_(rxn)=\sum n_(i).\Delta H_(f)(product)_(i)-\sum n_(j).\Delta H_(f)(reactant)_(j)`

where
`n_(i)` and
`n_(j)` represents number of moles of i-th product and j-th reactant in balanced reaction respectively.

Hence
`\Delta H_(rxn)=[1mol* \Delta H_(f)(CO_(2))_(g)]+[2mol* \Delta H_(f)(H_(2)O)_(l)]-[1mol* \Delta H_(f)(CH_(4))_(g)]-[2mol* \Delta H_(f)(O_(2))_(g)]`

so,
`\Delta H_(rxn)=[1mol* -393.5kJ/mol]+[2mol* -285.8kJ/mol]-[1mol* -74.8kJ/mol]+[2mol* 0kJ/mol]=-890.3 kJ`

So, Correct answer is -890.3 kJ