Methyl isocyanate, CH3-N-C-0, is used in the industrial synthesis of a type of pesticide and herbicide...

 

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Methyl isocyanate, CH3-N-C-0, is used in the industrial synthesis of a type of pesticide and herbicide known as a carbamate. As a historical note, an industrial accident in Bhopal, India, in 1984 resulted in leakage of an unknown quantity of this chemical into the air. An estimated 200,000 people were exposed to its vapors, and over 2000 of these people died. Reports from the site of this horrible accident indicate that the cause of death in the first wave of people who died was thermal burns to the lungs following inhalation of methyl isocyanate (H3C-N-C-0). Methyl isocyanate (MIC) reacts with water to form monoethylamine and carbon dioxide: H3C-N=C=0 + H₂O→ H3C-NH2 + CO₂ Estimate the amount of water, in moles, lining the lungs of someone your size (assume each lung has an inside surface area 4x the area of your hand, and a commensurate volume, and assume that the surface is covered with a film of water 0.1mm thick). Now estimate the amount of heat released inside the lungs of someone your size if 0.01, 0.05%, 0.10%, or 1.0% of that water reacted with MIC, and the corresponding increase in the temperature of the remaining water if all of the heat evolved in the reaction was absorbed by this water. Methyl isocyanate, CH3-N-C-0, is used in the industrial synthesis of a type of pesticide and herbicide known as a carbamate. As a historical note, an industrial accident in Bhopal, India, in 1984 resulted in leakage of an unknown quantity of this chemical into the air. An estimated 200,000 people were exposed to its vapors, and over 2000 of these people died. Reports from the site of this horrible accident indicate that the cause of death in the first wave of people who died was thermal burns to the lungs following inhalation of methyl isocyanate (H3C-N-C-0). Methyl isocyanate (MIC) reacts with water to form monoethylamine and carbon dioxide: H3C-N=C=0 + H₂O→ H3C-NH2 + CO₂ Estimate the amount of water, in moles, lining the lungs of someone your size (assume each lung has an inside surface area 4x the area of your hand, and a commensurate volume, and assume that the surface is covered with a film of water 0.1mm thick). Now estimate the amount of heat released inside the lungs of someone your size if 0.01, 0.05%, 0.10%, or 1.0% of that water reacted with MIC, and the corresponding increase in the temperature of the remaining water if all of the heat evolved in the reaction was absorbed by this water.