Question

Nitrosyl bromide (NOBr) can be obtained as a pure liquid at low temperatures. The liquid boils at –2°C, and at room temperature the gas partially decomposes, as shown below. 2 NOBr(g) Á 2 NO(g) + Br2(g) A 2.00-g sample of cold liquid NOBr is injected into a 1.0 L flask. (Assume that the flask was first evacuated so that it does not contain any air or other gases.) When the flask is allowed to come to equilibrium at 298 K, the total pressure inside is measured as 0.624 atm. (5 pts) (a). Calculate the total number of moles of gas present in the flask at equilibrium. (This part should be easy!) (b). Now find the numerical value of Keq for the reaction above at 298 K. (HINT: Set up the usual equilibrium table, and then try to relate the final concentrations to the total number of moles you found in part

Answer 1

a. Use gas law:    

n = pV/(RT) =

      = 0.624 atm x 1L /( 0.082 L.atmK^-1mol^-1 x 298K)

     = 0.0255 mol

b.

The initial quantity of NOBr was

2 g/ 109.9 g/mol = 0.0182 mol

2NOBr(g) = 2 NO(g) + Br2(g)

0.0182           0              0             initial state

0.0182-x         x             0.5 x        equilibrium state

0.0036        0.0146       0.0073     if x = 0.0146

The total number of moles at equilibrium is

0.0182-x + x   + 0.5 x = 0.0182 + 0.5 x = 0.0255 mol

X = 0.0146 mol

K = [NO]²[Br2]/[NO]²=

     = 0.0146² x 0.0073 / 0.0036² = 0.120