Question

find out the value of equilibrium constant for the following reaction at 298 Kelvin​

Answer 1

Answer:

Explanation:

USE THE RELATION Go=-2.30.RTlog(k)

Answer 2

The equilibrium constant for the given chemical reaction at 298 K is 239.9

Explanation:

For the given chemical equation:

$2NH_3(g)+CO_2(g)\rightleftharpoons NH_2CONH_2(aq.)+H_2O(l)$

To calculate the equilibrium constant for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-2.303RT\log K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -13.6 kJ/mol = -13600 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = 298 K

$K_{eq}$ = equilibrium constant at 298 K = ?

Putting values in above equation, we get:

$-13600J/mol=-(2.303\times 8.314J/Kmol)\times 298K\times \log K_{eq}\\\\K_{eq}=antilog (2.38)=239.9$

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