Home
Class 11
CHEMISTRY
Calculate the entropy change (J//mol K) ...

Calculate the entropy change `(J//mol K)` of the given reaction. The molar entropies `(J//K.mol)` are given in brackets after each substance:
`2PbS(s)[19.2]+3O_(2)(g)[205.1]`
` to 2PbO(s)[66.5]+2O_(2)(g)[248.2]`

A

`-113.5`

B

`-168.3`

C

`+72.5`

D

`-149.2`

Text Solution

Verified by Experts

The correct Answer is:
B

`Delta_(r )S=(2xx248.2)+(2xx66.5)`
`-(2xx91.2)-3(205.1)`
=-168.3 J
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|6 Videos
  • THERMODYNAMICS

    NARENDRA AWASTHI|Exercise Level 1 (Q.31 To Q.60)|2 Videos
  • STOICHIOMETRY

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|20 Videos

Similar Questions

Explore conceptually related problems

The decomposition of N_(2)O_(5) in C Cl_(4) at 318K has been studies by monitoring the concentration of N_(2)O_(5) in the solution. Initially the concentration of N_(2)O_(5) is 2.33 mol L^(-1) and after 184 minutes, it is reduced to 2.08 mol L^(-1). The reaction takes placed according to the equation 2N_(2)O_(5)(g)to 4 NO _(2)(g) +O_(2)(g) Calculate the average rate of this reaction in terms of hours, minutres and seconds. What is the rate of proudction of NO_(2) during this period ?

Calculate standard entropy change in the reaction Fe_(2)O_(3(s)) + 3H_(2(g)) rarr 2Fe_((s)) + 3H_(2)O_((l)) Given: S_(m)^(0) (Fe_(2)O_(3), s) = 87.4 S_(m)^(0) (Fe, s) = 27.3, S_(m)^(0) (H_(2)g) = 130.7, S_(m)^(0) (H_(2)O, 1) = 69.9 JK^(-1) mol^(-1)

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0^(@)C to ice at -10.0^(@)C . Delta_("fus")H=6.03 kJ mol^(-1) at 0^(@)C C_(p)[H_(2)O(l)]=75.3 J mol^(-1)K^(-1) C_(p)[H_(2)O(s)]=36.8 J mol^(-1)K^(-1)

Enthalpy of ammonia and water are -46.19KJ.mol^(-1) and -285.9KJ.mol^(-1) . Calculate AH for the reaction, 4NH_(3)(g)+3O_(2)(g)to 2N_(2)(g)+6H_(2)O(l) .

At the temperature T(K) for the reaction X_(2)O_(4) (l) rarr 2XO_(2)(g) Delta U = x kJ mol^(-1), Delta S = y JK^(-1) mol^(-1) . Gibbs energy change for the reaction is (Assume X_(2)O_(4), XO_(2) are ideal gases) Delta U = x kJ mol^(-1), Delta S = y JK^(-1) mol^(-1)

The temperature of K at which Delta G =0 for a given reaction with Delta H= -20.5 kJ mol^(-1) and Delta S =- 50.0 JK^(-1)mol^(-1)