26. At 32.4°C the reaction, Na SO4(s) + 1HNa So10 is at equilibrium and /-81.9 k

Question

26. At 32.4°C the reaction, Na SO4(s) + 1HNa So10 is at equilibrium and /-81.9 kJ. At the same temperature AS for (A) -268 J mol-K (B) 278 J mol K (C) 2527 J-mol-K (D) -2527 J mol K ( s reaction is 27. Two phases are considered to be in mutual equilibrium (A) both are at their critical temperatures. the mole fractions of all components are the same in each phase the temperature, a each phase. (B) (C) nent are sure, and chemical potential of each component are th (D) molecules are moving across the phase boundary + B C is rate =k(A)(Br.r[Alo = 1,0 M and [Bb-0.0010 M, then the rate law will be (A) pseudo-zeroth order. (B) pseudo-first order (C) pscudo-second order. (D) pseudo-third order. 29. Consider the contour diagram corresponding to the potential energy surface for the H+H: reaction when the three atoms (A. B and C) are constrained to be collinear. The point X on the diagram corresponds to the (A) steady state. (B) minimum energy state. (C) degenerate state. (D) transition state. 30. In heterogeneous catalysis of reactions on solid surfaces, the effectiveness of a given catalyst is not determined by the (A) bulk density of the catalyst. (B) surface roughness of the catalyst. (C) number of active sites on the catalyst surface. (D) degree of adsorption of the reactant on the catalyst surface. 3. The rate of chemical reactions depends on the total number of collisions between reactant molecules. The total number of collisions is favored by (A) smaller collision cross section and low molecular velocities. (B) smaller collision cross section and high molecular velocities. (C) larger collision cross section and low molecular velocities. (D) larger collision cross section and high molecular velocities.

Explanation / Answer

26)

deltaGo = 0.0 KJ/mol (At equilibrium, delta Go is 0)

deltaHo = -81.9 KJ/mol

T= 32.4 oC

= (32.4+273) K

= 305.4 K

we have below equation to be used:

deltaGo = deltaHo - T*deltaSo

0.0 = -81.9 - 305.4 *deltaSo

deltaSo = -0.2682 KJ/mol.K

deltaSo = -268 J/mol.K

Answer: A

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