2CO(g) + O2(g) 2CO2(g)
Go = -511.6 kJ and So = -173.1 J/K at 314 K and 1 atm.
The maximum amount of work that could be done by this reaction when 2.43 moles of CO(g) react at standard conditions at this temperature is__________ kJ.
how?!
To find the maximum amount of work that can be done by the reaction, we need to use the equation:
ΔG = ΔH - TΔS
where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin.
We are given the values for ΔG and ΔS. We can use these values to calculate ΔH using the equation:
ΔG = ΔH - TΔS
To rearrange the equation and solve for ΔH, we can add TΔS to both sides:
ΔH = ΔG + TΔS
Now we can substitute the given values into the equation:
ΔH = -511.6 kJ + (314 K)(-173.1 J/K)
First, convert the temperature to the same units as entropy:
314 K = 314 × 10^3 J
Now we can calculate:
ΔH = -511.6 kJ + (314 × 10^3 J)(-173.1 J/K)
ΔH = -511.6 kJ - 54.28 × 10^6 J^2/K
Next, we need to find the maximum work done by the reaction. The maximum work is equal to -ΔG, which is the negative of the change in Gibbs free energy. So:
Maximum work = -ΔG
Plugging in the given value for ΔG:
Maximum work = -(-511.6 kJ)
Maximum work = 511.6 kJ
Therefore, the maximum amount of work that could be done by this reaction when 2.43 moles of CO(g) react at standard conditions at this temperature is 511.6 kJ.