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chap 7/7
| Question | Answer |
|---|---|
| WHY IS ENDO REACTION FASTER... | COZ THE EA IN THE ENDO DIRECTION IS GREATER IN THE EXOTHERMIC REACTIONS WITH HIGH EA ARE MORE SENSITIVE TO TEMP CHANGES THAN REACTION WITH LOWER EA. |
| REACTIONS WITH HIGH EA ARE MORE... | SENSITIVE TO TEMP CHANGES THAN REACTION WITH LOWER EA. |
| AFFECT OF PRESSURE ONLY APPLIES TO EQUILIBRIUM... | REACTIONS IN INVOLVING GASES. |
| PRESSURE CAUSED BY.... | BOMBARDMENT OF GAS MOLECULES ON WALLS OF CONTAINER. |
| A GIVEN TEMP PRESSURE DEPENDS ON ... | THE NUMBER OF GAS MOLECULES IN A GIVEN VOLUME. |
| PRESSURE OF A EQUILIBRIUM SYSTEM CAN BE INCREASED BY.... | REDUCING VOLUME, 2. ADDING MORE GAS INTO THE SAME VOLUME. MORE MOLECULES PER CM3. |
| THE DIRECTION OF CHANGE OF EQUILIBRIUM POS DEPENDS ON.... | NO OF GAS MOLECULE (OR MOLES OF GAS) ON EACH SIDE OF EQUATION. |
| IF PRESSURE DECREASED LE CHATELIER’S PRINCIPLE PREDICTS ... | THAT THE SYSTEM WILL REACT IN ORDER TO TRY BRING THE PRESSURE DOWN AGAIN TOWARDS THE ORIGINAL VALUE. |
| CAN ONLY BRING PRESSURE DOWN IF .... | EQUILIBRIUM SHIFTS TO SIDE WITH FEWER GAS MOLECULES. |
| INCREASES IN PRESSURE (AT CONSTANT TEMP) WILL CAUSE... | POSITION OF EQUILIBRIUM TO SHIFT TOWARDS SIDE OF EQUATION WITH FEWER GAS MOLECULES (FEWER MOLES OF GAS). |
| PREDICT OF INCREASING THE PRESSURE IN EQUATION: N2(G) + 3H2(G) REVERSE ARROWS 2NH3(G) CHANGE IN ENTHALPY -92.4KJ ... | THERE ARE 4 MOLECULES ON THE LEFT SIDE OF EQUATION AND ONLY 2 ON THE RIGHT. CAUSE POSITION TO SHIFT TO SIDE WITH FEWER GAS MOLECULES. WHICH IS TO THE RIGHT. |
| HIGHER PRESSURES WILL INCREASE THE PROPORTION OF .... | NITROGEN AND HYDROGEN THAT IS CONVERTED INTO AMMONIA. |
| PREDICT INCREASING THE PRESSURE IN : H2(H) +I2(G) REVERSE ARROWS 2HI(G) ENTHALPY CHANGE=+51.8KJ. .. | 2 MOLES OF GAS ON EACH SIDE. INCREASE PRESSURE WILL HAVE NO EFFECT ON POSITION OF EQUILIBRIUM. |
| METHANE CAN BE TRAPPED IN WATER AS .... | A SOLID METHANE HYDRATE (CH4(H2O)6]. |
| REACTION BETWEEN METHANE GAS AND WATER TO FORM METHANE HYDRATE IS... | EXOTHERMIC. CH4(G0 + 6H20(L) REVERES ARROW [CH4(H20)6](S) |
| METHANE HYDRATE FOUND AT... | BOTTOM OF SEA OFF CANADIAN COAST WHERE PRESSURE HIGH. |
| PREDICT IF PRESSURE ON SOLID METHANE HYDRATE REDUCED TO 1ATM... | SHIFT TO THE SIDE WITH MORE GAS MOLECULES. THIS MEANS METHANE HYDRATE WILL DECOMPOSE AND METHANE GAS WILL BE PRODUCED. |
| PREDICT WHAT WILL HAPPEN IF METHANE HYDRAE WAS HEATED... | REACTION IS EXOTHERMIC. INCREASE IN TEMP WILL CAUSE EQUILIBRIUM POSITION TO SHIFT IN THE ENDOTHERMIC DIRECTION/TO THE LEFT. THUS METHANE HYDRATE WILL DECOMPOSE. |
| PREDICT OF METHANE HYDRATE CRYSTALS WHERE BOUGHT TO THE SURFACE OF THE SEA? ... | THE SOLID METHANE HYDRATE FROM COLD OCEAN FLOOR MEANS A DECREASES IN PRESSURE AND INCREASES IN TEMPERATURE. THUS RESULTING IN METHANE HYDRATE DECOMPOSING INTO METHANE GAS AND WATER. |
| AFFECT ON CONC ONLY APPLIES TO SYSTEM ... | WHERE THE REACTANT ARE DISSOLVED IN SOLVENT E.G. WATER |
| IF THE CONC SUBS IN EQUILIBRIUM REACTION INCREASED, LE CHATELIER’S PREDICTS WILL... | REMOVE SOME OF THAT SUBSTANCE |
| TEST FOR BROMIDE AGBR(S) +2NH3(AQ) REVERES ARROWS [AG (NH3)2]+(AQ) +BR-(AQ)THE POS OF EQ.. | IS NORMALLY TO THE LEFT, SO VERY LITTLE SILVER BROMIDE DISSOLVES IN DILUTE AQUEOUS AMMONIA. |
| HOWEVER IF CONCENTRATED AMMONIA IS ADDED .... | [NH3] INCREASES. DRIVES THE POSITION OF EQ TO THE RIGHT, |
| CAUSING THE PRECIPITATE OF SILVER BROMIDE TO DISAPPEAR AS REACTS. |
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