33467
In the reversible reaction \(A + B\) \(\rightleftharpoons\) \(C + D\), the concentration of each \(C\) and \(D\) at equilibrium was \(0.8\) mole/litre, then the equilibrium constant \({K_c}\) will be
1 \(6.4\)
2 \(0.64\)
3 \(1.6\)
4 \(16\)
Explanation:
(d) Suppose \(1\) mole of \(A\) and \(B\) each taken then \(0.8\) mole/litre of \(C\) and \(D\) each formed remaining concentration of \(A\) and \(B\) will be \((1 -0.8) = 0.2\) mole/litre each. \(Kc = \frac{{[C]\,\,[D]}}{{[A]\,\,[B]}} = \frac{{0.8 \times 0.8}}{{0.2 \times 0.2}} = 16.0\)
Chemical Equilibrium
33468
\(4\) moles of \(A\) are mixed with \(4\) moles of \(B\). At equilibrium for the reaction \(A + B\) \(\rightleftharpoons\) \(C + D\), \(2 \) moles of \(C\) and \(D\) are formed. The equilibrium constant for the reaction will be
33469
On a given condition, the equilibrium concentration of \(HI,\,{H_2}\) and \({I_2}\) are \(0.80, \,0.10\) and \(0.10\) mole/litre. The equilibrium constant for the reaction \({H_2} + {I_2}\) \(\rightleftharpoons\) \(2HI\) will be
33467
In the reversible reaction \(A + B\) \(\rightleftharpoons\) \(C + D\), the concentration of each \(C\) and \(D\) at equilibrium was \(0.8\) mole/litre, then the equilibrium constant \({K_c}\) will be
1 \(6.4\)
2 \(0.64\)
3 \(1.6\)
4 \(16\)
Explanation:
(d) Suppose \(1\) mole of \(A\) and \(B\) each taken then \(0.8\) mole/litre of \(C\) and \(D\) each formed remaining concentration of \(A\) and \(B\) will be \((1 -0.8) = 0.2\) mole/litre each. \(Kc = \frac{{[C]\,\,[D]}}{{[A]\,\,[B]}} = \frac{{0.8 \times 0.8}}{{0.2 \times 0.2}} = 16.0\)
Chemical Equilibrium
33468
\(4\) moles of \(A\) are mixed with \(4\) moles of \(B\). At equilibrium for the reaction \(A + B\) \(\rightleftharpoons\) \(C + D\), \(2 \) moles of \(C\) and \(D\) are formed. The equilibrium constant for the reaction will be
33469
On a given condition, the equilibrium concentration of \(HI,\,{H_2}\) and \({I_2}\) are \(0.80, \,0.10\) and \(0.10\) mole/litre. The equilibrium constant for the reaction \({H_2} + {I_2}\) \(\rightleftharpoons\) \(2HI\) will be
33467
In the reversible reaction \(A + B\) \(\rightleftharpoons\) \(C + D\), the concentration of each \(C\) and \(D\) at equilibrium was \(0.8\) mole/litre, then the equilibrium constant \({K_c}\) will be
1 \(6.4\)
2 \(0.64\)
3 \(1.6\)
4 \(16\)
Explanation:
(d) Suppose \(1\) mole of \(A\) and \(B\) each taken then \(0.8\) mole/litre of \(C\) and \(D\) each formed remaining concentration of \(A\) and \(B\) will be \((1 -0.8) = 0.2\) mole/litre each. \(Kc = \frac{{[C]\,\,[D]}}{{[A]\,\,[B]}} = \frac{{0.8 \times 0.8}}{{0.2 \times 0.2}} = 16.0\)
Chemical Equilibrium
33468
\(4\) moles of \(A\) are mixed with \(4\) moles of \(B\). At equilibrium for the reaction \(A + B\) \(\rightleftharpoons\) \(C + D\), \(2 \) moles of \(C\) and \(D\) are formed. The equilibrium constant for the reaction will be
33469
On a given condition, the equilibrium concentration of \(HI,\,{H_2}\) and \({I_2}\) are \(0.80, \,0.10\) and \(0.10\) mole/litre. The equilibrium constant for the reaction \({H_2} + {I_2}\) \(\rightleftharpoons\) \(2HI\) will be
33467
In the reversible reaction \(A + B\) \(\rightleftharpoons\) \(C + D\), the concentration of each \(C\) and \(D\) at equilibrium was \(0.8\) mole/litre, then the equilibrium constant \({K_c}\) will be
1 \(6.4\)
2 \(0.64\)
3 \(1.6\)
4 \(16\)
Explanation:
(d) Suppose \(1\) mole of \(A\) and \(B\) each taken then \(0.8\) mole/litre of \(C\) and \(D\) each formed remaining concentration of \(A\) and \(B\) will be \((1 -0.8) = 0.2\) mole/litre each. \(Kc = \frac{{[C]\,\,[D]}}{{[A]\,\,[B]}} = \frac{{0.8 \times 0.8}}{{0.2 \times 0.2}} = 16.0\)
Chemical Equilibrium
33468
\(4\) moles of \(A\) are mixed with \(4\) moles of \(B\). At equilibrium for the reaction \(A + B\) \(\rightleftharpoons\) \(C + D\), \(2 \) moles of \(C\) and \(D\) are formed. The equilibrium constant for the reaction will be
33469
On a given condition, the equilibrium concentration of \(HI,\,{H_2}\) and \({I_2}\) are \(0.80, \,0.10\) and \(0.10\) mole/litre. The equilibrium constant for the reaction \({H_2} + {I_2}\) \(\rightleftharpoons\) \(2HI\) will be
