Question

The enthalpy of formation for C6 H6 (I) is 49.0 kJ/mol. Consider the following reaction.

6 upper C (s, graphite) plus 3 upper H subscript 2 (g) right arrow upper C subscript 6 upper H subscript g (l).

Is the reaction endothermic or exothermic, and what is the enthalpy of reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
exothermic; Delta.Hrxn = 49.0 kJ
exothermic; Delta.Hrxn = –49.0 kJ
endothermic; Delta.Hrxn = 49.0 kJ
endothermic; Delta.Hrxn = –49.0 kJ

Answer 1

Answer:

Using the formula, we subtract the enthalpy of the reactants from that of the product. The reactants have an enthalpy of 0. Thus:

ΔH = 49 - 0

ΔH = 49 kJ/mol

Explanation:

Hope this helps.

Answer 2

Answer:

What is Delta.Hrxn for this reaction?

⇒ -2803 kJ

Explanation: