During a spontaneous chemical reaction, it is found that δSsys < 0. This means __________

a. δSsurr < 0 and its magnitude is < δSsys.
b. δSsurr < 0 and its magnitude is > δSsys.
c. δSsurr > 0 and its magnitude is < δSsys.
d. δSsurr > 0 and its magnitude is > δSsys.
e. an error has been made, as Ssys > 0 by necessity for a spontaneous process.
Heat

e is the answer, if S is entropy

To answer this question, we need to understand the concept of entropy (S) in the context of chemical reactions.

Entropy is a measure of the disorder or randomness in a system. In a spontaneous chemical reaction, the total entropy change of the universe (ΔSuniv) must be greater than zero. This means that the sum of the entropy change in the system (ΔSsys) and the entropy change in the surroundings (ΔSsurr) must be positive:

ΔSuniv = ΔSsys + ΔSsurr > 0

Now, the question states that ΔSsys is less than zero (δSsys < 0) in a spontaneous reaction. This means that the entropy of the system decreases during the reaction, indicating a decrease in disorder or randomness.

To satisfy the equation above and ensure that ΔSuniv is positive, the magnitude of ΔSsurr must be greater than δSsys. This is because the negative value of ΔSsys needs to be compensated by a larger positive value of ΔSsurr to yield a positive total entropy change.

Therefore, the correct answer is option (d) - δSsurr > 0 and its magnitude is > δSsys.