Question

Consider the reaction 6CO2(g) + 6H2O(I)-->C6H12O6 + 602(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.36 moles of CO2(g) react at standard conditions. DELTA S surroundings=__ J/K

Answer 1

The entropy change for the surroundings when 2.36 moles of CO2(g) react at standard conditions is -1281.87 J/K.

Step-by-step explanation:

The entropy change for the surroundings can be calculated using the equation:

ΔS surroundings = -ΔH/T

where ΔH is the enthalpy change for the reaction and T is the temperature in Kelvin. To find ΔH, we need to multiply the coefficient of CO2 by the enthalpy change of CO2 (which is given as -393.5 kJ/mol) and multiply the coefficient of water by the enthalpy change of water (which is given as -286 kJ/mol). Then, we can sum up these values to find the total enthalpy change for the reaction. Finally, we can plug these values into the equation to calculate ΔS surroundings.

ΔS surroundings = (-393.5 kJ/mol * 6 + -286 kJ/mol * 6) / 298 K

ΔS surroundings = -1281.87 J/K