A medium can be anything that a wave can pass through e.g the medium can be solid objects, air, water, etc.
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
The second one is the way to go.
Answer : The amount of heat evolved by a reaction is, 4.81 kJ
Explanation :
Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water
![q=[q_1+q_2]](https://tex.z-dn.net/?f=q%3D%5Bq_1%2Bq_2%5D)
![q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]](https://tex.z-dn.net/?f=q%3D%5Bc_1%5Ctimes%20%5CDelta%20T%2Bm_2%5Ctimes%20c_2%5Ctimes%20%5CDelta%20T%5D)
where,
q = heat released by the reaction
= heat absorbed by the calorimeter
= heat absorbed by the water
= specific heat of calorimeter = 
= specific heat of water = 
= mass of water = 254 g
= change in temperature = 
Now put all the given values in the above formula, we get:
![q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]](https://tex.z-dn.net/?f=q%3D%5B%28783J%2F%5EoC%5Ctimes%20-2.28%5EoC%29%2B%28254g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20-2.28%5EoC%29%5D)
Therefore, the amount of heat evolved by a reaction is, 4.81 kJ
The answer is B . Brønsted-Lowry Acid/bases trade H+
