Question

Using the information provided calculate the heat of reaction involved in the conversion of 25.00 g of methane (CH4, MM=16.0426 g/mol) to chloroform (CHCl3, MM=119.3779 g/mol):

CH4(g) + 3 Cl2(g) --> CHCl3(l) + 3 HCl(g)

Given the following:
C(graphite) + 2 H2(g) --> CH4(g)
deltaH = -92.3 kJ

C(graphite) + 1/2 H2(g) + 3/2 Cl2(g) --> CHCl3(l)
dH = -74.5 kJ

1/2 H2(g) + 1/2 Cl2(g) --> HCl(g)
dH = -135.1 kJ

ANS:
-603.9 kJ

Much help is appreciated! Thank you!

Answer 1

Reverse equn 1 (and the sign of dH)
add in equn 2 as is.
Multiply eqn 3 by 3 and add in. Multiply the dH by 3 also.
You should have dH rxn of 92.3 - 74.5 - 3(-135.2) = -?
Then dHrxn x (25.0/molar mass CH4) = ?

Answer 2

Thank you very much!