Consider the reaction H2CO(g) + O2(g)CO2(g) + H2O(l) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.63 moles of H2CO(g) react at standard conditions. S°surroundings = J/K

3,116J/K

Explanation:

This question asks to calculate the entropy change of the surroundings.

To do this, we need the standard enthalpy of formation ΔfH° of the reacting species and products first:

We should observe that standard enthalpy if formation of O2 is zero. We proceed with the rest of the species.

H2CO = -109.5KJ/mol

CO2 = -393.5KJ/mol

H2O = -285.8KJ/mol

Now, we calculate the standard change of enthalpy of the reaction as:

ΔHrxn = ΔHproduct - ΔHreactant = (-285.8 - 393.5) +(109.5) = -569.8 KJ/mol

The relationship between the entropy and the standard formation enthalpy is given as

The relationship is:

ΔSosurroundings = - ( ΔHof/ T)

We convert the standard enthalpy of formation to joules first = -569.8 * 10^3 Joules

Using the formula above at a temperature of 298k, the entropy change would be:

-(-569.8 * 10^3)/298 = 1912J/K

Now, we know that 1.63 moles of H2CO reacted. We also need to know the coefficient of the H2CO in the reaction which is 1.

We thus have:

1.63 mol H2CO(g) * (1912J/K * 1 mol H2CO) = 3116J/K

smooth muscle is composed of involuntary muscles found within the walls of organs and structures such as the esophagus, stomach, intestines, and blood vessels. those muscles push materials like food or blood through organs. unlike skeletal muscle, smooth muscle can never be under your control.