25K Find the K. or K, values for each of the reactions.K2 K-2.5x10 at 25C K,-.02

Question

25K Find the K. or K, values for each of the reactions.K2 K-2.5x10 at 25C K,-.0288 at 45 C Consider the reaction: No. (g) # 2 NO2(g) taking place in a sealed, fixed volume container. Suppose initially [??. (g)1-0 and [NO2(g) -3 0 M. Further suppose that at some temperature the reaction proceeds to equilibrium and [NO(g. 2.0 M,i.e the equilibrium concentration of [NO:(g)]. -2.0 M. Sketch a reasonable plot of molar concentrations of each substance as a function of time showing the progress of reaction to reach equilibrium. (Make sure the axes are labeled!! Make sure you show linear increments along each axis!!). 8. 3 3 ts2

Explanation / Answer

Answer:

(6) Given reaction is 2SO2 (g) + O2 (g) <------> 2SO3 (g)

Kc=2.5x10-2 at 25°C

The realation between Kc and Kp is

Kp=Kc(RT)?n

Where R=gas constant (0.0821 bar•litre/mol•K), T is temperature =25°C=25+273K=298 K,

and ?n=Sum of moles of gaseous products ?- Sum of moles of gaseous reactants)=2-(2+1)=-1

Therefore Kp=(2.5x10-2) (0.0821 bar•litre/mol•K x 298 K)-1

Kp=(2.5x10-2)/(0.0821 bar•litre/mol•K x 298 K)

Kp=1.02x10-3.

(7) Given reaction 3N2 (g) + 9H2 (g) <-----> 6NH3 (g)

Kp=0.0288 at 45°C

The realation between Kc and Kp is

Kp=Kc(RT)?n

Where R=gas constant (0.0821 bar•litre/mol•K), T is temperature =45°C=45+273K=318 K,

and ?n=Sum of moles of gaseous products ?- Sum of moles of gaseous reactants)=6-(9+3)=-6

Therefore (0.0288)=Kc (0.0821 bar•litre/mol•K x 318 K)-6

Kc=(0.0288) x (0.0821 bar•litre/mol•K x 318 K)6

Kc=9.12 x 106.

As per guidelines I have to answer first question. But I answerd two for you.

Thanks and I hope you like it.

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