Consider the reaction 2na(s) + 2h2o(l)2naoh(aq) + h2(g) using standard thermodynamic data at 298k, calculate the entropy change for the surroundings when 1.74 moles of na(s) react at standard conditions. s°surroundings = j/k g

Answer : The entropy change for the surroundings when 1.74 moles of Na(s) react at standard conditions is

Explanation :

The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction .

where,

We are given:

Putting values in above equation, we get:

Now we have to calculate the entropy change for surrounding .

where,

= change in entropy

= change in enthalpy = -652.5 kJ

T = temperature = 298 K

Now put all the given values in the above formula, we get:

Now we have to calculate the entropy change for the surroundings when 1.74 moles of Na(s) react at standard conditions.

As, 2 moles of Na(s) has entropy change =

And, 1.74 moles of Na(s) has entropy change =

Thus, 23 g of Na(s) has entropy change =

Therefore, the entropy change for the surroundings when 1.74 moles of Na(s) react at standard conditions is

c is the answer because when something is left after something evaporates that shows the chemical change.