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please solve 6.5 and 6.4 sum... Given the reaction : A 63 mole Of A and one mole of B 1.0 litre vessel. After established, 0.4 mole of C is present in 64 The equilibrium constant for the Of co-a reaction, + H20 (g) CO at a certain temperature is 2.2. Initially one mole 6•10 of CO and one mole of H20 are placed in a 2.0L acid is container. What are the concentrations of all the substances when the reaction reaches equilibrium? ([C02] = [H2] = 0.2986M, = = (1.83 ) 6.12 0.2014M).

6.5 Calculate Kp for the reaction, c(s) + H20 (g) CO(g) + H2(g) 6.13 at 990 K if the equilibrium concentrations are as 6.14 ca follows : [H20] = 1.10M, [COI = = 0.2M, R = 0.08206 L atm mor . 6.15 (Hint : First calculate Kc) (Kp 2.954) 6.6 The equilibrium constant Kc for the reaction —-—-.y 2NH3 (g) is 0.286 at 5000C. Calculate Kc and Kp for the 6.1t

The solution for Q.no:6.5 is as expressed below, (kindly post each question separately for quick answering)

$q.no:6.5\phantom{\rule{0ex}{0ex}}C\left(s\right)+{H}_{2}O\left(g\right)\iff CO\left(g\right)+{H}_{2}\left(g\right)\phantom{\rule{0ex}{0ex}}{K}_{c}=\frac{\left[CO\right]\left[{H}_{2}\right]}{\left[{H}_{2}O\right]}=\frac{0.2M\times 0.2\overline{)M}}{1.10\overline{)M}}=0.03636M=0.03636\frac{mol}{L}\phantom{\rule{0ex}{0ex}}{K}_{P}={K}_{C}(RT{)}^{\u2206n}=0.03636\frac{\overline{)mol}}{\overline{)L}}(0.08206\frac{\overline{)L}atm}{\overline{)K}\overline{)mol}}\times 990\overline{)K}{)}^{1}=2.95atm$

Regards.

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