How much heat would be required to convert 5.31 mol of a pure substance from a liquid at 40.0°C to a gas at 113.0°C?

Q = 7.7 KJ

Explanation:

The complete question carries the following data:

Specific Heat Capacity of Liquid State = C(l) = 1.45 J/mol.°C

Specific Heat Capacity of Gaseous State = C(g) = 0.65 J/mol.°C

Boiling Temperature = Tb = 88.5°C

Heat of Vaporization = ΔH(vap) = 1.23 KJ/mol

Now, first we calculate the heat required (Q₁) to raise he temperature to boiling point:

Q₁ = n*C(l)*ΔT₁

where,

n = no. of moles = 5.31 mol

ΔT₁ = Temperature difference from 40°C to Tb = 88.5°C - 40°C = 48.5°C

Therefore,

Q₁ = (5.31 mol)(1.45 J/mol.°C)(48.5°C)

Q₁ = 373.4 J = 0.37 KJ

Now, we calculate the heat required (Q₂) to change its phase:

Q₂ = nΔH(vap)

Q₂ = (5.31 mol)(1.23 KJ/mol)

Q₂ = 6.53 KJ

Now, we calculate the heat required (Q₃) to raise he temperature to boiling point:

Q₃ = n*C(g)*ΔT₃

where,

n = no. of moles = 5.31 mol

ΔT₃ = Temperature difference from Tb to 113°C = 113°C - 88.5°C = 24.5°C

Therefore,

Q₃ = (5.31 mol)(0.65 J/mol.°C)(24.5°C)

Q₃ = 84.5 J = 0.08 KJ

So, the total heat required to convert 5.31 mol of this pure substance from a liquid at 40°C to a gas at 113°C is:

Q = Q₁ + Q₂ + Q₃

Q = 0.37 KJ + 6.53 KJ + 0.08 KJ

Q = 7.7 KJ

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2