Given the following, determine Delta G degree f at 298 K for SnO. Sn(s) + SnO_2(

Question

Given the following, determine Delta G degree f at 298 K for SnO. Sn(s) + SnO_2(s); Delta G degree = 12.0 kJ at 298 K A) -251.9 kJ/mol B) -503.8 kJ/mol C) 527.8 kJ/mol D) 263.9 kJ/mol E) 1055.6 kJ/mol The following reaction is spontaneous at all temperatures: CaC_2(s) + 2H_2O(l) rightarrow Ca(OH)_2(s) + C_2H_2(g) The following reaction is spontaneous at all temperatures: CaC_2(s) + 2H_2O(l) rightarrow Ca(OH)_2(s) + C_2H_2(g) Which of the following statements is true? A) Delta H is negative and AS is positive. B) Delta H is negative and Delta S is negative. C) Delta H is positive and Delta S is negative. D) Delta H is positive and Delta S is positive. E) Delta G is positive at all temperatures. ____ Which of the three laws of thermodynamics provides a criterion for spontaneity? A) the third law of thermodynamics B) the first law of thermodynamics C) the second law of thermodynamics D) both the second and third laws of thermodynamics For the reaction below Delta G degree = + 33.0 kJ, Delta H degree = 92.2 kJ, and Delta S degree = +198.7 J./K. Estimate the temperature at which this reaction becomes spontaneous. 2 NH_3(g) rightarrow N_2(g) + 3 H_2(g) A) 0.464 K B) 298 K C) 464 K D) 166 K

Explanation / Answer

For the reaction, deltaG= 2* delta G of SnO- { 1*deltaG of Sn + 1*deltaG of SnO2}= 12

2, 1,1 are coefficients of SnO, Sn and SnO2 repsectively

= 2* x-1*0+515.8= 12 , since deltag of sn=0 and let that of SnO=x

2x= 12-515.8, x=-251.9 Kj/mole ( A is correct)

2. for a reaction to be spontaneous. deltaG= deltaH-T*delta S need to be -ve

for deltaG to be negative, deltaH has to be -ve and deltaS has to be +ve. this makes deltaG -ve. A is correct.

3. Second law of thermodynamics disucses that for any process to be spontaneous. detlaS>0. first law discusses about law of conservation of energy and third law discusses about entropy but not spontanity. C is correct

4. deltaG= deltaH-T*deltaS

fora reaction to be spontaneous, deltaG <0

so let us calculate at what temperature deltaG =0

92.2- T*197.7/1000 =0

T=464K

T>464 K the reaction becomes spontaneous. ( C is correct)

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