For the reversible reaction,$A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,$\Delta G^{\circ} = -350 \ kJ$,which one of the following statements is true?
- AThe reaction is thermodynamically non-feasible
- BThe entropy change is negative
- CEquilibrium constant is greater than one
- DThe reaction should be instantaneous
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Similar Questions
Calculate $\Delta G^\circ$ for the conversion of oxygen to ozone $\frac{3}{2} O_{2(g)} \to O_{3(g)}$ at $298 \ K$,if $K_p$ for this conversion is $2.47 \times 10^{-29}$.
Medium
View Solution$STATEMENT-1$: For every chemical reaction at equilibrium,standard Gibbs energy of reaction is zero. $STATEMENT-2$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.
Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$
Difficult
View SolutionThe equilibrium constant $(K)$ of a reaction may be written as :
The equilibrium constant of the following given reaction is $K_p = 6.022 \times 10^{-5}$ at $298 \ K$ temperature. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. Calculate the value of $\Delta_r G^o$.
Medium
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