-0.145 kj
Any colour change.
The reaction which forms potassium hydrogen carbonate from potassium carbonate, K2CO3 + H2O + CO2 --> KHCO3 is difficult to perform in a laboratory; the same goes for the decomposition. Hence instead of simply using a calorimeter or similar apparatus, it is necessary to use another reaction route and a Hess Cycle using two reactions such as: reaction A: K2CO3 + 2HCl --> 2KCL + H2O +CO2 reaction B: KHCO3 + 2HCl --> KCl + H2O +CO2 The enthalpy change of the decomposition of potassium hydrogen carbonate will be twice the enthalpy change for reaction B, minus the enthalpy change for reaction A.
it is because potassium hydroxide will react with carbon dioxide to form potassium carbonate and water. but potassium carbonate is soluble, so there will be no visible change to see if there is carbon dioxide emitted
Yes, the reaction involving the solid is actually an individual step in the equation of the reaction between the solutions. If you were to add the change in enthalpy of the reaction with the solid NaOh to the change in enthalpy of the other step in the reaction (that's adding water and the NaOh pellets) you would find the sum equivalent to the change in enthalpy of the reaction involving the two solutions (this is supported by Hess's law). I suggest that you consider Hess's law for more information.
Chemical Change
Any colour change.
The reaction which forms potassium hydrogen carbonate from potassium carbonate, K2CO3 + H2O + CO2 --> KHCO3 is difficult to perform in a laboratory; the same goes for the decomposition. Hence instead of simply using a calorimeter or similar apparatus, it is necessary to use another reaction route and a Hess Cycle using two reactions such as: reaction A: K2CO3 + 2HCl --> 2KCL + H2O +CO2 reaction B: KHCO3 + 2HCl --> KCl + H2O +CO2 The enthalpy change of the decomposition of potassium hydrogen carbonate will be twice the enthalpy change for reaction B, minus the enthalpy change for reaction A.
it is because potassium hydroxide will react with carbon dioxide to form potassium carbonate and water. but potassium carbonate is soluble, so there will be no visible change to see if there is carbon dioxide emitted
Chemical Change
Yes, the reaction involving the solid is actually an individual step in the equation of the reaction between the solutions. If you were to add the change in enthalpy of the reaction with the solid NaOh to the change in enthalpy of the other step in the reaction (that's adding water and the NaOh pellets) you would find the sum equivalent to the change in enthalpy of the reaction involving the two solutions (this is supported by Hess's law). I suggest that you consider Hess's law for more information.
enthalpy change of solution=enthalpy change of hydration - enthalpy change of lattice
Enthalpy change of neutralisation is defined as the enthalpy change of a reaction where one mole of hydrogen ions reacts with one mole of hydroxide ions to form one mole of water under standard conditions of 1 atm, 298K (25 degree Celsius) and in the solutions containing 1 mol per dm3.
Yes - you have an acid and base and the resultant products are a salt (Potassium Chloride) and water
It can be argued it does as Potassium is a silvery gray metal. When reacted with water it produces colorless Potassium Hydroxide and colorless hydrogen gas.
Enthalpy is the energy absorbed or lost from a reaction, but enthalpy change per mole is the amount of energy lost per mole, so in order to get the overall enthalpy from the change per mole, you must multiply that value by the amount of moles used in the reaction.
The presence of a catalyst affect the enthalpy change of a reaction is that catalysts do not alter the enthalpy change of a reaction. Catalysts only change the activation energy which starts the reaction.
It will change colour and become purple