Question

Calculate ΔSsurr for a reaction for which ΔHrxn= - 1575 kJ at 25°C. Assume constant temperature and pressure

a. -529 x 10³ J/K
b. 4.69 x 10⁵J/K
c. 6.30 x 10⁹ J/K
d. 5.29 x 10³ J/K

Answer 1

The ΔSsurr for a reaction with ΔHrxn = -1575 kJ at 25°C is approximately 5.29 x 103 J/K.

Step-by-step explanation:

To calculate the change in entropy of the surroundings (ΔSsurr) for a reaction with a given change in enthalpy (ΔHrxn), at a constant temperature, we use the formula:

ΔSsurr = -ΔHrxn / T

Given that ΔHrxn = -1575 kJ and the temperature T is 25°C (or 298 K), we convert ΔHrxn to joules since we need consistent units:

• ΔHrxn = -1575 kJ * 1000 J/kJ = -1575000 J

Now, we plug the values into the formula:

ΔSsurr = -(-1575000 J) / 298 K

ΔSsurr = 5283.22 J/K

Therefore, the ΔSsurr for the reaction at 25°C is approximately 5.29 x 103 J/K, which corresponds to option d.