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Anika [276]
3 years ago
15

What is the value of k for this aqueous reaction at 298 k?

Chemistry
1 answer:
TEA [102]3 years ago
5 0

Answer : The complete question is attached in answer.

The value of K will be = 2 X10^{-5}


Explanation : We can use the formula as;


∆G = -RT ln K


on rearranging we get, K = e^({-deltaG/RT})

Therefore, we get,

∆G/RT = (26.81 kJ/mol ) / (0.008314 kJ/mol-K) X (298K) = 10.82


So, K = e^{-10.82}

Therefore, K = 2 X 10^{-5}

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What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?
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Answer : The value of reaction quotient, Q is 0.0625.

Solution : Given,

Concentration of N_2 = 2.00 M

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The balanced equilibrium reaction is,

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The expression of reaction quotient for this reaction is,

Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}

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3 years ago
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