enthalpy of combustion kj mol
H2 + 1/2 o2 = h20 + 286 kJ
how much did you see? 62.4 ml H2
1 mol of any gas at NTP = 24.5 L
You need to clarify your question. Also define NTP.
I apologize for any confusion. NTP stands for Normal Temperature and Pressure, which is a standardized set of conditions used in gas calculations. In this case, NTP refers to a temperature of 25 degrees Celsius (298.15 Kelvin) and a pressure of 1 atmosphere (atm).
To determine the enthalpy of combustion in kJ/mol, we can use the given balanced equation:
H2 + 1/2 O2 → H2O + 286 kJ/mol
Now let's say we have 62.4 mL of H2 gas. To convert this volume to moles, we can use the conversion factor:
1 mol of any gas at NTP = 24.5 L
First, convert the volume from mL to L:
62.4 mL ÷ 1000 = 0.0624 L
Now, we can determine the number of moles of H2 gas:
Number of moles = volume (in liters) / 24.5 L/mol
Number of moles = 0.0624 L / 24.5 L/mol
Number of moles = 0.00255 mol
Therefore, we have 0.00255 moles of H2 gas. Now, we can determine the enthalpy of combustion:
Enthalpy of combustion = Number of moles of H2 gas × Enthalpy change per mole
Enthalpy of combustion = 0.00255 mol × 286 kJ/mol
Enthalpy of combustion = 0.7303 kJ (rounded to four decimal places)
Therefore, the enthalpy of combustion for 62.4 mL of H2 gas is approximately 0.7303 kJ.