The combustion of pentane, c5h12, occurs via the reaction c5h12(g)+8o2(g)→5co2(g)+6h2o(g) with heat of formation values given by the following table: substance δh∘f (kj/mol) c5h12 (g) -119.9 co2(g) −393.5 h2o(g) −241.8 calculate the enthalpy for the combustion of pentane. express your answer to four significant figures and include the appropriate units.

 -3298.4 kJ/mol

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

where,

n = number of moles

(as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

Therefore, the enthalpy change for this reaction is, -3298 kJ/mol

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

if you mean hydrogen and oxygen then water is the product