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

each chemical family is represented as a group or column on the periodic table. based on chemical properties, there are 6 types of chemical families. group 1 represents alkali metals, group 2 represents alkaline earth metals, groups 3 to 12 represent transition metals, groups 13 to 16 represent non-metals and other metals, group 17 represents halogens and group 18 represents noble gases.