Consider the following reaction in a closed reaction flask. If 1.50 atm of gas A

Question

Consider the following reaction in a closed  reaction flask.  If 1.50 atm of gas A is allowed to react with 0.500 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. (0.250 atm of A3B produced) + (0.500 atm of A remaining) = 0.750 atm total pressure B. (0.500 atm of A3B produced) + (0.500 atm of A remaining) = 1.00 atm total pressure C. 0.500 atm of A3B is produced, and since no other gases remain, 0.500 atm is the total pressure D. (0.500 atm of A3B produced) + (1.00 atm of A remaining) = 1.50 atm total pressure

Explanation / Answer

6A(g) + 2B(g) ? 2A3B(g)

as per the abovw eq. for 1mole of A 1/3 moles of B is required to form 1/3 moles of A3B

0.500 atm of A3B is produced, and since no other gases remain, 0.500 atm is the total pressure

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