- 05 15 Pts The Rate Law For The Following Second Order Reaction At 10 C Can Be Written As 2 Nobr 2 No G Bra 1 (26.68 KiB) Viewed 30 times
05. (15 pts) The rate law for the following second-order reaction at 10 °C can be written as: 2 NOBr () ► 2 NO(g) + Bra
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05. (15 pts) The rate law for the following second-order reaction at 10 °C can be written as: 2 NOBr () ► 2 NO(g) + Bra
05. (15 pts) The rate law for the following second-order reaction at 10 °C can be written as: 2 NOBr () ► 2 NO(g) + Bra (8) Reaction # 1 2 3 [AB] 0.2 0.4 0.6 Rate (mol/L) 3.2 x10 1.28 x 102 2.88 x 102 1/2 Rate for above reaction is k[NOBr] where k = 0.80 L/mol s. Show your work to receive full credit. (a) Determine thet when (NOBrlo = 0.650M. (b) Calculate [NOBr) at t=5.80 x 10's if [NOBr)o = 0.650M. (c) If [NOBr]o = 1.00M, how long would it take for 50 % NOBr to react?
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