enthalpy change^H is +23kJ mol-1 of phosphine (PH3) for the reaction

2P(s)red+3H2(g)---->2PH3(g)
a) Calc. entropy change for this reaction. Explain how this information allows you to determine the spontaneity or non-spontaneity of this reaction at any temp (S for PH3 is 210JK-1 mol-1)
b) do these results indicate anything about the possibility of the existence of phosphine? explain your answer.

a) To calculate the entropy change (∆S) for the reaction, you need to consider the stoichiometric coefficients of the balanced chemical equation and the standard entropy values of the reactants and products. The equation for calculating ∆S is:

∆S = ΣnS(products) - ΣnS(reactants)

Where:
- ΣnS(products) is the sum of the standard entropy values (S) of the products, multiplied by their respective stoichiometric coefficients (n).
- ΣnS(reactants) is the sum of the standard entropy values (S) of the reactants, multiplied by their respective stoichiometric coefficients (n).

Given the equation:
2P(s)red + 3H2(g) → 2PH3(g)

The standard entropy value for PH3 is 210 J K^(-1) mol^(-1). Since there are two PH3 molecules formed in the reaction, the contribution from the products would be:
2 × 210 J K^(-1) mol^(-1) = 420 J K^(-1) mol^(-1)

For the reactants, there are 2 moles of P(s)red and 3 moles of H2(g) involved. However, solid P does not have a standard entropy value since it is in its standard state, so its entropy contribution is considered zero. The hydrogen gas (H2) does have a standard entropy value, but you have not provided that information.

To evaluate the spontaneity of the reaction, you need to compare the calculated entropy change (∆S) with the entropy change calculated from the formula:
∆G = ∆H - T∆S

Where:
∆G is the change in Gibbs free energy,
∆H is the enthalpy change, and
T is the temperature in Kelvin.

If ∆G < 0, the reaction is spontaneous at that temperature. If ∆G > 0, the reaction is non-spontaneous at that temperature.

b) The existence of phosphine (PH3) is indicated by the positive enthalpy change (∆H) of +23 kJ mol^(-1). Since the enthalpy change is positive, it means that energy is required for the reaction to occur, leading to the formation of PH3. This suggests that phosphine can exist under the given reaction conditions.