Terrorist Chemical Releases: Assessment of Medical Risk and Implications for Emergency Preparedness

ABSTRACT

In the last few years, public health has played an increasingly important role in disaster management, particularly in biological terrorist event planning. However, little time or financial expenditure has been spent on preparation for terrorist-related chemical events. In addition, chemical hazardous material and industrial accidents are common occurrences in the United States and have significant public health and emergency preparedness consequences. This manuscript reviews previous terrorist-related and industrial chemical events, an assessment of the risk that these events have on the public health, and ways that hospitals and local, state, and regional public health agencies may plan for such an event.

Key Words:

• emergency preparedness
• chemical agents
• terrorism
• risk assessment
• emergency medicine.

Notes

¹ LD50 = lethal dose 50%: This is the dose of liquid that would kill 50% of people to whom it was applied.

² LCT50 = lethal concentration-time 50%: This is the concentration of airborne chemical that would kill 50% of people who breathed it for one minute.

³ Volume of stadium = 1,172,127 ft² × 215 ft (height of roof) = 130236 m² × 66 m = 8,595,576 m³. Kg sarin required = LCT50 × Vol = 100 mg min/m³ × 8,595,576 m³ = 860 Kg min. Liters sarin required = 860 kg × 1 mL/1.1 g (at 20C) (Sidell et al. 1997, p. 141) = 781L.

^4. Arena volume = 383,528 ft² × 150 ft height = 57,529,200 ft³ = 35,649 m² × 46 m = 1,639,854 m³. Kg sarin required = LCT50 × Vol = 100 mg min/m³ × 1,639,854 m³ = 164 Kg min. Liters sarin required = 164 kg × 1 mL/1.1 g (at 20C) (Sidell et al. 1997, p. 141) = 149 L.

^5. Ct50(miosis): This is the concentration at which 50% of exposed patients would develop miosis after exposure for one minute (Sidell et al. 1997). Volume required in baseball stadium = Ct50 × volume of stadium × 1 mL/1.1 g sarin = 3 mg min/m³ × 8,595,576 m³ × 0.91 mL/g = 23.4 L = 6.2 gallons. Volume required in hockey arena = Ct50 × volume of arena × 1 mL/1.1g sarin = 3 mg min/m³ × 1,639,854 m³ × 0.91 mL/g = 4.48 L = 1.2 gallons.