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S(H(2(g)))^(0) = 130.6 JK^(-1) mol^(-1),...

`S_(H_(2(g)))^(0) = 130.6 JK^(-1) mol^(-1), S_(H_(2)O_((l)))^(0) = 69.9JK^(-1) mol^(-1), S_(O_(2(g)))^(0) = 205 JK^(-1)`, Then the absolute entropy change of `H_(2(g)) + (1)/(2)O_(2(g)) rarr H_(2)O_((l))` is

A

`-163.2 Jmol^(-1) K^(-1)`

B

`+ 163.2 J mol^(-1) K^(-1)`

C

`-303 J mol^(-1) K^(-1)`

D

`+303J mol^(-1) K^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta S= S_(P) - S_(R ) = S_(H_(2)O)^(0) - S_(H_(2))^(0) - (1)/(2) S_(O_(2))^(0)`
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