Calculate the S values for the following reactions. In eachcase explain the sign

Question

Calculate the S values for the following reactions. In eachcase explain the sign of S.

a) N2H4(s) +H2(g) -->2NH3(g)

N2H4(s) = 238.5 J/mol-K
H2(g) = 130.58 J/mol-K
NH3(g) = 192.5 J/mol-K


b) 2Al(s) + 3Cl2(g) -->2AlCl3(s)

Al(s) = 28.32 J/mol-K
Cl2(g) = 222.96 J/mol-K
AlCl3(s) = 109.3 J/mol-K


c) Mg(OH)2(s) + 2HCl(g) -->MgCl2(s) + 2H2O(l)

Mg(OH)2(s) = 63.24 J/mol-K
HCl(g) = 186.69 J/mol-K
MgCl2(s) = 89.6 J/mol-K
H2O(l) = 69.91 J/mol-K


d) 2CH4(g) -->C2H6(g) +H2(g)

CH4(g) = 186.3 J/mol-K
C2H6(g) = 229.5 J/mol-K
H2(g) = 130.58 J/mol-K


THANK YOU!

Explanation / Answer

For each reaction's S value you have to use theequation Sreaction =Sproducts - Sreactants Add the S values of the products together and add the Svalues of the reactants, then subtract the reactants S fromthe products. a) For NH3 remember to multiply the S value by 2because of its coefficient in the reaction. Using the process abovethe answer is: 15.92 J/mol-K. The sign ofthe total reaction's S is positive, so that means the amountof disorder is increasing. b) Once again, remember the coefficients...The answer is -506.9J/mol-K. The sign of the totalS is negative so that means order is increasing. c) Same process as above: The answer is -207.2J/mol-K. The sign of S is negative so the productsare more orderly than the reactants. d) The answer is -12.52 J/mol-K. S isnegative so order is increasing.

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