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What is equilibrium expression for the reaction $ P_4 + 5O_2 \rightleftharpoons P_4O_{10} $
$ P_{4(S)}+ 5O_{2(g)} \rightleftharpoons P_4O_{10(s)} $ $ K_C = { [P_4 O_{10} \over [P_{4(s) ] [ O_{2(g)] ^ 5 $ we know that concentration of a solid component is always taken as a unity $ \therefore Kc = { 1 \over [ O_2 ] ^ 5 $
Partial pressure of $ O_2$ in $ 2Ag_2 O(g) \rightleftharpoons 4 Ag(s) + O_{2 (g) } $ is
$ 2Ag_2O_{(s)} \rightleftharpoons 4 Ag_{(s)} + O_{2_{(g)} } $ for this reaction $ Kp = Po_2 $
For reaction $ H_{2(g)} + CO_{2(g)} \rightleftarrow CO_{(g)} + H_2O(g) ,If the initial concentrationof [H_2] = [CO_2] and x moles / litre of hydrogen is consumed at equilibrium, the correct expression of K_P $ is
$ H_{2(g) } + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2 O_{(g)}$ initial
conc. 1 1 0 0 At eqm. (1-x) (1-x) x x $ Kp = { P_{CO} \times P_{H_2O} \over P_{H_2} \times P_{CO_2} } ={ x^2 \over (1-x)^2 $
Consider the imaginary equlibrium $ 4A + 5B \rightleftharpoons 4x + 6y $ The equilibrium constant $ K_C $ has unit
For reaction $ CO_{(g) } + { 1 /2 } O_{2(g)} \rightleftharpoons CO_{2(g)} { Kp \over Kc } $ is equivalent to
For the reaction $ CH_{4(g)} + 2 O_{2(g)} \rightleftharpoons CO_{2(g)} + 2 H_2 O_{(L)} \triangle H = - 170.8 KJ mol ^ {-1} $ Which of following statement is not true
For the given reaction \( CH_{4(g)} + 2 O_{2(g)} \rightleftharpoons CO_{2(g)} + 2 H_2O_{(L)} \Delta H = -170.8 \, KJ \, mol^{-1} \), we need to determine which statement is not true. The equilibrium constant expression for a reaction involving gases is typically given in terms of partial pressures or concentrations of the gaseous reactants and products. However, in this case, \( H_2O \) is in the liquid state, so its concentration is taken as constant and omitted from the equilibrium expression. Thus, the equilibrium constant expression should be: \[ Kp = \frac{[CO_2]}{[CH_4][O_2]^2} \] Therefore, option o4 is not true.
The reaction Quetient (Q) for the reaction $ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_3 $ is given by $ Q = { [ NH_3] ^2 \over [N_2] [H_2 ] ^ 3 } $ The reaction will proceed from right to left is
The reaction quotient (Q) gives the ratio of the concentrations of products to reactants at any point in time. For the reaction \( N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_3 \), \[ Q = \frac{[NH_3]^2}{[N_2][H_2]^3} \] The direction in which the reaction will proceed depends on the comparison between Q and the equilibrium constant \( K_c \). If \( Q > K_c \), the reaction will proceed in the reverse direction (from right to left) to reach equilibrium. Therefore, the correct option is \( Q > K_c \).
Which of following is not favourable for formating $ SO_3 $ formation $ 2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)} \triangle H = -45. 0 Kcal $
The formation of $SO_3$ from $SO_2$ and $O_2$ is an exothermic reaction, as indicated by the negative enthalpy change ($ riangle H = -45.0 ext{ Kcal}$). According to Le Chatelier's Principle, increasing the temperature would shift the equilibrium to the left, favoring the reactants and thus not favoring the formation of $SO_3$. Therefore, high temperature is not favorable for $SO_3$ formation.
The most important buffer in blood consists of
The most important buffer system in blood consists of carbonic acid ($H_2CO_3$) and bicarbonate ($HCO_3^-$). This buffer system maintains the pH of blood by neutralizing acids and bases, ensuring that the pH remains within the narrow range necessary for proper physiological function.
Select the pKa value of strongest acid from following
The strength of an acid is inversely related to its pKa value; the lower the pKa value, the stronger the acid. Among the given options, a pKa value of 1 is the lowest, indicating that it corresponds to the strongest acid.
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