Question

question 4 (1 point)
when 6.000 moles of glucose are combusted, 16 815 kj of energy is produced. which of the following chemical equations represent this process?
6c6h12o6(s) + 36o2(g) ® 36co2(g) + 36h2o(ℓ) dh = -2802.15 kj
6c6h12o6(s) + 36o2(g) ® 36co2(g) + 36h2o(ℓ) dh = 16815 kj
c6h12o6(s) + 6o2(g) ® 6co2(g) + 6h2o(ℓ) dh = -16815 kj
c6h12o6(s) + 6o2(g) ® 6co2(g) + 6h2o(ℓ) dh = -2802.15 kj
all equations are correct

Explanation:

Step1: Calculate energy per mole

Given \(6.000\) moles of glucose produce \(16815\) kJ. Energy per mole \(=\frac{16815}{6}=2802.15\) kJ. Combustion is exothermic (\(\Delta H<0\)).

Step2: Analyze equations

• For \(6C_{6}H_{12}O_{6}(s)+36O_{2}(g)

ightarrow36CO_{2}(g)+36H_{2}O(l)\), \(\Delta H=- 2802.15\times6=-16815\) kJ (incorrect coefficient - energy relation).

• For \(C_{6}H_{12}O_{6}(s)+6O_{2}(g)

ightarrow6CO_{2}(g)+6H_{2}O(l)\), \(\Delta H=-2802.15\) kJ (correct as per per - mole calculation).

Answer:

\(C_{6}H_{12}O_{6}(s)+6O_{2}(g)
ightarrow6CO_{2}(g)+6H_{2}O(l)\), \(\Delta H=-2802.15\) kJ (the fourth option)