Seeking help with physical chemistry problems~~
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A chemical reaction takes place in an adiabatic container with a constant volume of 0.5 m3. The temperature of the gas inside the container increases by 750 degrees Celsius, and the pressure rises by 600 KPa. What is the change in enthalpy for this process?
There should be two methods: 1. Determine the H values before and after the reaction separately and subtract them; 2. Calculate based on the standard reaction enthalpy change, by first converting the state before the reaction to the standard state, carrying out the reaction, and then converting the state back to that after the reaction, and finally taking the algebraic sum of the H values for these processes.
According to the first law of thermodynamics, U=Q-W and H=U+PV; for an adiabatic container, the change in enthalpy should be zero
Adiabatic means there is no heat exchange with the outside environment, but the reaction itself releases heat, causing the temperature to rise; therefore, the enthalpy change is not zero
Adiabatic constant volume is equivalent to an isolated system; therefore, its thermal energy remains unchanged, that is, its internal energy stays constant. According to the definition of enthalpy, H = U + pv, so the change in enthalpy is equal to pv times the change in value, which is 0.5 multiplied by 600 = 300 kJ
dH = dU + dPV = dU + PdV + VdP. Here, dU = dQ – dW = 0 – 0 = 0. Since it is a constant-volume process, PdV = 0; therefore, dH = VdP. For an ideal gas, using PV = nRT, we have dH = (nRT/P)dP. Upon integration, ΔH = nRT*lnP, where nRT can be replaced with P1V1