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3) The reaction below has a Kc value of 3.64 x 10-3. What is the value of Kp for this reaction at 25°C? 2 NaN3(s) - 2 Na(s) +
5) Determine the value of Ke for the following reaction if the equilibrium concentrations are as follows: [N2 Jeq - 3.6 M, [O
0 0
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Answer #1

Kp=ke (RT) An An= 3 R = 0.082) T = 298 K Kp= 3.640203 (0.0821x298) Kp = 53.2 (ND)? 102) (6.612 (4.1)

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