Question

3) The complete combustion of hexane has a H of -8326kJ/mol for the balanced reaction

a. Write the balanced equation for the combustion of pentane

b. What is the enthalpy change for the reverse reaction?

c. Balance the equation in the forward direction with whole number coefficients

d. What is the enthalpy change for the reaction in problem “c”?

e. Which reaction is more likely to be thermodynamically favored; the forward or reverse?

f. Explain your answer to e.

g. If the reaction was written to produce H2O gas instead of liquid, would the magnitude of H increase, decrease or stay the same?

h. Explain your answer to g.

Answer 1

A.  $C5H12+ O2\rightarrow H2O+C O2$ by balancing we get ,

$C5H12+8 O2\rightarrow6 H2O+5C O2$

B. Enthalpy is the heat content of the system at constant pressure. It is denoted by H. The absolute value of enthalpy cannot be determined.Enthalpy of a system is a state function, thus the magnitude of the enthalpy change depends only upon the enthalpies in the initial and final states i.e   $\Delta H= Hfinal-Hinitial$

For any reation ,enthalpy change of the forward reaction and reverse reaction will be equal in magnitude and has opposite signs. For eg: enthalpy change of a forward reaction is -834 kj and enthalpy change for a reverse reaction must be +834 kj

c. we can take the reaction, #+9 —> H, O here H₂ to Ho Had H=2 ) CD 0=2 0=1 deactant side a hydrogen product side a hydrogen and 2 oxygen and losygen. we can add a in the front of bydrogen in the product section, so equation becomes, Hz +9—52403 1433 H=2x24 (2) 02 0=x222 hence, the oxygens molecule in both sides balanced ire [0=2 021x2=2) bext, we have to again add a in the front of hydrogens in the creactant so, equations (2) becomes 2H₂ +2 -> 2H2O H2 2x2=4 Hz 2x2=4 022 0 21x2=2 firsally equation balanced - 24 to 240