the following reaction is carried out in a bomb calorimeter: 4NH3 (g) +5O2 (g)-->4NO (g)+6H2O (g)
The calorimeter constant is 45.0kJ/K. when 34.0g of NH3 are reacted in the calorimeter, the temp rises by 10 degrees C. What is the change in energy, delta E, for the rxn?"
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To find the change in energy (ΔE) for the reaction, we can use the equation:
ΔE = q - CΔT
Where:
- ΔE is the change in energy (in this case, the energy released or absorbed by the reaction)
- q is the heat absorbed by the bomb calorimeter (in this case, the heat absorbed by the reaction)
- C is the calorimeter constant
- ΔT is the change in temperature in Kelvin
First, let's convert the mass of NH3 from grams to moles. The molar mass of NH3 is approximately 17.03 g/mol.
34.0 g NH3 * (1 mol NH3 / 17.03 g NH3) = approximately 1.997 mol NH3
Now, let's calculate the heat absorbed by the bomb calorimeter (q). This can be determined by the equation:
q = CΔT
Given:
- C (calorimeter constant) = 45.0 kJ/K
- ΔT (change in temperature) = 10 °C
However, to use the equation, we need to convert ΔT from Celsius to Kelvin by adding 273.15.
ΔT (Kelvin) = 10 °C + 273.15 = 283.15 K
Now, let's calculate q:
q = 45.0 kJ/K * 283.15 K
q ≈ 12716.75 kJ
Finally, we can find the change in energy (ΔE) by substituting the calculated values into the equation:
ΔE = q - CΔT
ΔE = 12716.75 kJ - 45.0 kJ/K * 283.15 K
Calculating the multiplication part:
ΔE = 12716.75 kJ - 12741.75 kJ
ΔE ≈ -25.0 kJ
Therefore, the change in energy (ΔE) for the reaction is approximately -25.0 kJ.