At 500°C the equilibrium constant, K p , is 4.00 × 10 -4 for the equilibrium:...

Question

Question

At 500°C the equilibrium constant, K p , is 4.00 × 10 -4 for the equilibrium: 2HCN( g) <--> H 2( g) + C 2N 2( g )

What is K p for the following reaction? H 2 ( g ) + C 2 N 2 ( g ) <--> 2HCN( g ).

science chemistry

Answer 1

Answer #1

Answer 2

Similar Homework Help Questions

At 500°C the equilibrium constant, Kp , is 4.00 ×10–4 for the equilibrium: 2HCN(g) ⇌ H2(g)...

At 500°C the equilibrium constant, Kp , is 4.00 ×10–4 for the equilibrium: 2HCN(g) ⇌ H2(g) + C2N2(g) What is Kp for the following equilibrium? 4HCN(g) ⇌ 2H2(g) + 2C2N2(g)
The equilibrium constant, K. for the following reaction is 1.20 times 10^2 at 500 K An...

The equilibrium constant, K. for the following reaction is 1.20 times 10^2 at 500 K An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.166 M PCl_5 4.47 times 10^-2 MPCl_5 and 4.47 times 10^-2 M Cl_2 What will be the concentrations of the three gases once equilibrium has been re-established if 3.26 times 10^-2 mol of Cl_2(g) is two gases once equilibrium has been reestablished? [NH_3} = M [H_2S] = M
The equilibrium constant, K, for the following reaction is 1.20 times 10^-2 at 500 k. PCl_5...

The equilibrium constant, K, for the following reaction is 1.20 times 10^-2 at 500 k. PCl_5 (g) irreversible PCl_3 (g) + Cl_2 (g) At equilibrium mixture of the three gases in a 1.00 L. flask at 500 K contains 0.203 M PCl_5, 4.93 times 10^-2 M PCl_5 and 4.93 times 10^-2 M Cl_2. What will be the concentration of the three gases once equilibrium has been reestablished, if 2.81 times 10^-2 mol of PCl_3 (g) is added to the flask?...

The equilibrium constant, K., for the following reaction is 1.20 10-2 at 500 K. PCI(8) =PC13(g)...

The equilibrium constant, K., for the following reaction is 1.20 10-2 at 500 K. PCI(8) =PC13(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.363 moles of PC13() are introduced into a 1.00 L vessel at 500 K M [PC131 = [PC13] - [Cl] M M
1. Write down the equilibrium constant expressions, K, and K for each of the following reactions:...

1. Write down the equilibrium constant expressions, K, and K for each of the following reactions: (a) H2(g)C(g) 2 HCl(g) (b) 2 C(s)+0(g) 2 COg Ag (aq)Cl(aq) (c) AgCl(s) (d) 2 0,(g) 30,(g) 2. A 1.0 L evacuated flask was charged with 0.020 mol of N,O and 0.060 mol of NO, at 25.0 C. After equilibrium was reached the NO2 concentration was found to be 0.0140 M. What is the equilibrium constant K, for the reaction? N,0,(g)2 NO(g) 3. Ammonium...
The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) +...

The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.287 moles of PCl5(g) are introduced into a 1.00 L vessel at 500 K. [ PCl5] = M [PCl3] = M [Cl2] = M

1) The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g)...

1) The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.311 M HI, 4.18×10-2 M H2 and 4.18×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.85×10-2 mol of I2(g) is added to the flask? 2) The equilibrium constant, K, for the following reaction is 1.20×10-2 at...
13) The value of the equilibrium constant K, for the formation of ammonia, N, (g) +...

13) The value of the equilibrium constant K, for the formation of ammonia, N, (g) + 3 H, () € 2 NH, (e) is 4.75 x 10 at 450 K. What is the K, for the following reaction 2 N N(g) + 3 H; (@)
The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PCl3(g) + Cl2(g)...

The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PCl3(g) + Cl2(g) = PCl5(g) Calculate the equilibrium concentrations of reactant and products when 0.249 moles of PCl3 and 0.249 moles of Cl2 are introduced into a 1.00 L vessel at 500 K. [PCl3] = M [Cl2] = M [PCl5] = M The equilibrium constant, Kc, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) = H2(g) + I2(g) Calculate the equilibrium concentrations of reactant and...

1. The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) <---->...

1. The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K: PCl5(g) <----> PCl3(g) + Cl2(g) Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.00 atm at 500 K. PPCl5 = atm PPCl3 = atm PCl2 = atm 2. The equilibrium constant, Kp, for the following reaction is 0.215 at 673 K: NH4I(s) <----> NH3(g) + HI(g) Calculate the equilibrium partial pressure of HI when...