Consider the following reaction in a closed reaction flask. If 1.00 atm of gas A
ID: 794812 • Letter: C
Question
Consider the following reaction in a closed reaction flask. If 1.00 atm of gas A is allowed to react with 1.00 atm of gas B and the reaction goes to completion at constant temperature and volume, what is the total pressure in the reaction flask at the end of the reaction? What are the partial pressures of the gases in the flask?
6A(g) + 2B(g) ? 2A3B(g)
A. (0.667 atm of A3B produced) + (0.333 atm of B remaining) = 1.00 atm total pressure B. (0.333 atm of A3B produced) + (0.667 atm of B remaining) = 1.00 atm total pressure C. (1.00 atm of A3B produced) + (0.333 atm of A remaining) = 1.33 atm total pressure D. (0.333 atm of A3B produced) + (0.167 atm of A remaining) = 0.500 atm total pressureExplanation / Answer
intial moles of A be A0
intial moles of B be B0
6A(g) + 2B(g) ? 2 A3B(g)
let volume be V and temperature be T
we have P V = n R T
A0 = (1 * V)/(R*T) = V/( R * T )
B0 = (1 * V)/(R*T) = V/( R * T )
in final state we have
Af = 0 , Bf = (2 * V )/(3 * R * T ) , product (A3B) = (1 * V )/(3 * R * T )
final total pressure in the flask = (R T Bf )/(V) + (R *T *A3Bf )/(V) = (2/3) + (1/3) = 1atm
partial pressures of the gases in the flask are
partial pressure for B = (R T Bf )/(V) = 2/3 = 0.667
partial pressure for A3B = (R * T *A3Bf )/(V) = 1/3 = 0.333
(0.333 atm of A3B produced) + (0.667 atm of B remaining) = 1.00 atm total pressure