(1) Derive a rate law assuming surface reaction is rate-limiting. (10%)

1.One hundred moles per minute of a binary mixture of 50% A and 50% B are separated in a two-stage (serial) process. In the first stage, the liquid and vapor flow rates exiting from the stage are each 50 mol/min. The liquid stream is then passed into a second separator that operates at the same temperature as the first stage. The temperature is the same for each stage, and at that temperature, the vapor pressure of A is 10 kPa and the vapor pressure of B is 100 kPa. Treat the liquids and vapors as ideal. Calculate the compositions of the vapor and liquid streams, and the pressure in the first stage. (30%)

(1) Derive an expression relating, to measurable properties. (10%)

(2) Find for the gas in this process. (10%) ＜Hint＞ Expansion rule:

3. Consider a second-order reaction 2A→B+C, which is occurred in 20 meters of schedule 40 pipe packed with catalyst. There is 104.4 Ibm/h of gas flowing through the bed with the entering pressure of 101300 Pa. Temperature was kept at 260℃. Entering volumetric flow rate (vo) is 6.5 m3 /h. Cross-sectional area of pipe is 0.0013 m2. Catalyst pellet size is 6 mm； catalyst bulk density is 1800 kg/m3. Pressure drop parameter (β0) is 2580 Pa/m. Assume specific reaction rate constant (k) is independent of particle size； k = 0.012 m6/mol kg cat. h. Calculate the conversion with and without pressure drop. (20%) If the catalyst size increases by a factor of 2, what is the conversion with pressure drop? (10%)

(2) Derive a rate law assuming adsorption of X is rate-limiting. Let Y adsorb on catalyst surface and react with Xㆍ S. (10%) ＜Hint＞ Modified ( B ) and add one extra adsorption reaction.

(一)證明正斷層受到重力滑動的功 。

(二)對正斷層滑動的功 W 進行一次微分，以決定斷層滑動的最佳傾角。

(三)對正斷層滑動的功 W 進行二次微分，以判定上述最佳斷層傾角為最小重力滑動的功所造成。

(一)為何建築物固有角頻率為 ？

(二)求建築物水平位移隨時間的變化 x(t)為何？