Question

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the following equation: CO2(s)→CO2(g). When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 liters of water heated to 90 ∘C.

Part A

Use standard enthalpies of formation to calculate the change in enthalpy for dry ice sublimation. (The ΔH∘f for CO2(s) is - 427.4kJ/mol ).

Part B

Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 14 ∘C . Assume no heat loss to the surroundings.

Answer 1

part A

CO2(s)→CO2(g)

DHrxn = DH0fCO2(g) - DH0fCO2(S)

             = (-393.5) - (-427.4)    = 33.9 kj/mol

part B

heat lost by warm water (q) = m*s*DT

                mass of water(m) =   V*D   =   15*10^3*1 = 15*10^3g

            S = specific heat of water =4.18 j/g.c

    DT = 90- 14 = 76 C

q = 15*10^3*4.18*76

      = 4765.2 kj

DHrxn =   33.9 kj/mol

that means 1 mole CO2(s) = 44 g CO2(s) consumes = 33.9 kj

amount of CO2(s) required = 4765.2/33.9*44 = 6184.92 g

                   = 6.2 kg