Answer:
The equation to show the the correct form to show the standard molar enthalpy of formation:
Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Given, that 1 mole of 
gas and 1 mole of 
liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.
Divide the equation by 2.
The equation to show the the correct form to show the standard molar enthalpy of formation:
Answer:
Explanation:
Scientific laws or laws of science are statements, based on repeated experiments or observations, that describe or predict a range of natural phenomena.[1] The term law has diverse usage in many cases (approximate, accurate, broad, or narrow) across all fields of natural science (physics, chemistry, astronomy, geoscience, biology). Laws are developed from data and can be further developed through mathematics; in all cases they are directly or indirectly based on empirical evidence. It is generally understood that they implicitly reflect, though they do not explicitly assert, causal relationships fundamental to reality, and are discovered rather than invented.[2]
Answer:
I hope this is it. I'm not really sure.
Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
