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 : ![2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol](/tex.php?f=2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol)
equation 3 : ![4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol](/tex.php?f=4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol)
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 : ![4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol](/tex.php?f=4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol)
let us add equation 1 & 4
![4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol](/tex.php?f=4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol)
![4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol](/tex.php?f=4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol)
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