In a chemical reaction,the rate constant for the backward reaction is $7.5 \times 10^{-4}$ and the equilibrium constant is $1.5$. The rate constant for the forward reaction will be .......
- A$2 \times 10^{-3}$
- B$5 \times 10^{-4}$
- C$1.12 \times 10^{-3}$
- D$9.0 \times 10^{-4}$
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The value of $\log \ K$ for the reaction $A \rightleftharpoons B$ at $298 \ K$ is (Nearest integer).
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)
Given: $\Delta H^{\circ} = -54.07 \ kJ \ mol^{-1}$
$\Delta S^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$
(Take $2.303 \times 8.314 \times 298 = 5705$)
When one mole of $A$ and one mole of $B$ were heated in a one-litre flask at $T \ K$,$0.5 \ mole$ of $C$ was formed at equilibrium for the reaction $A + B \rightleftharpoons C + D$. The equilibrium constant,$K_C$,is:
For the reaction,$CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$,the ratio $K_p/K_c$ is equal to:
MediumAIEEE 2004
View SolutionFor the reaction $C_{(s)} + CO_{2(g)} \rightleftharpoons 2CO_{(g)}$,at equilibrium,$P_{CO_2} = 2$ and $P_{CO} = 4$. The value of $K_p$ is:
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View SolutionIn the thermal dissociation of $PCl_5$,the total pressure in the gaseous equilibrium mixture is $1.0 \ atm$ when half of $PCl_5$ is found to dissociate. The equilibrium constant of the reaction $(K_p)$ in atmosphere is
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