Calculate the reaction enthalpy δrh when 1 mol of glucose is oxidized to acetic acid at 80°c. 2o2(g)+c6h12o6(s)→2ch3cooh(l)+2co2( g)+2h2o(l) express your answer to four significant figures and include the appropriate units.

The reaction enthalpy for the oxidation of 1 mole of glucose is -1,055 kJ.

Explanation:

The standard enthalpy of a reaction can be calculated from the standard enthalpies of formation of each substance, using the following expression:

ΔH°r = ∑n(p).ΔH°f(p) - ∑n(p).ΔH°f(p)

where,

ΔH°r is the standard enthalpy of the reaction

n(i) is the number of  moles of reactants and products in the balanced equation

ΔH°f(i) is the standard enthalpy of formation of reactants and products, which are tabulated.

Considering the balanced equation:

2 O₂(g) + C₆H₁₂O₆(s) → 2 CH₃COOH(l) +2 CO₂(g) +2 H₂O(l)

The standard enthalpy of reaction is:

ΔH°r = [2 x ΔH°f(CH₃COOH(l)) + 2 x ΔH°f(CO₂(g)) + 2 x ΔH°f(H₂O(l))] - [2 x ΔH°f(O₂(g)) + 1 x ΔH°f(C₆H₁₂O₆(s))]

ΔH°r = [2 mol x (-483.5 kJ/mol) + 2 mol x (-393.5 kJ/mol) + 2 mol x (-285.8 kJ/mol)] - [2 mol x 0 kJ/mol + 1mol x (-1,271 kJ/mol)]

ΔH°r = -1,055 kJ

In the balanced equation there is 1 mole of glucose. So, the enthalpy of reaction for such amount is -1,055 kJ (exothermic).

the answer is   c.) 1575 j.

explanation:

14 days is the half-life of phosphorus-32