A) CH 4 (g) + 4Cl 2 (g) > CCl 4(g) + 4HCl(g) Hf CH 4 (g) = -74.9 kJ/mol, Hf Cl 2(g) = 0 kJ/mol, Hf CCl 4(g) = -106.6 kJ/mol,
Hf HCl (g) = - 92.3kJ/mol

b) CH 4(g) + 2O 2(g) > CO 2(g) + 2 H2 O(l)
Hf CH 4 (g) = -74.9 kJ/mol, H f O 2 (g) = 0 kJ/mol, H f CO 2 (g) = -393.5 kJ/mol,
Hf H 2 O (l) = - 285.5kJ/mol

c) 4FeS 2(s) + 11O 2(g) 2 Fe 2 O 3(s) + 8 SO 2(g)
H f FeS 2(s) = -177.5 kJ/mol, H f O 2 (g) = 0 kJ/mol, H f Fe 2 O 3(s) = -822.2 kJ/mol,
H f SO 2(g) = - 296.8kJ/mol

d) 3 C (s) + 4 H 2(g) C 3 H 8 (g)
C 3 H 8 (g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O(l ) H c C 3 H 8(g) = - 2219.9kJ/mol

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol