Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.87 x 10³ at
-
- Site Admin
- Posts: 899603
- Joined: Mon Aug 02, 2021 8:13 am
Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.87 x 10³ at
5.00 + x < PREV 3 NEXT > Based on the given values and your value for Qc, set up ICE table in order to determine the unknown. Where applicable, use the x variables to represent any unknown change in concentration. Initial (M) Change (M) Equilibrium (M) Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.87 x 10-³ at a certain temperature. If 5.00 g of solid PH3BCI, and 0.0700 g of BCI, are added to a 4.500 L reaction vessel, what will the equilibrium concentration of PH, be? PH₂BCI3(S) PH3(g) + BCl3(g) 0 5.00 - x 2 PH,BCI, (s) 5.00 0.700 0.0700+ x 0.0700 - x PH₂(g) 4 + 8.27 x 10-3 1.33 x 10-4 BCI, (g) +x RESET -X 8.27 x 10+. 8.27 x 10- 1.33 x 10+ 1.33 x 104 - x
[x] [8.27 x 10+ [2x] Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.87 x 10³ at a certain temperature. If 5.00 g of solid PH,BCI, and 0.0700 g of BCI, are added to a 4.500 L reaction vessel, what will the equilibrium concentration of PH, be? < PREV 2 NEXT > Based on your ICE table, set up the expression for Kc in order to determine the unknown. Do not combine or simplify terms. Kc = [8.27 x 10- PH,BCI, (s) PH3(g) + BCl3(g) 1 [x]² - [1.33 x 10+ 4 = 1.87 x 10-³ [2x]¹ [5.00+ x] (1.33 x 10-x (8.27 x 10+ [5.00 -x] [0.0700+x] (8.27 x 10- (1.33 x 10+ RESET [0.0700 -x] (1.33 x 10-x
Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.87 x 10³ at a certain temperature. If 5.00 g of solid PH,BCI, and 0.0700 g of BCI, are added to a 4.500 L reaction vessel, what will the equilibrium concentration of PH, be? PH₂BCI3(S) PH3(g) + BCl3(g) < PREV 1 2 Based on your ICE table and expression for Kc, solve for the partial pressure of PH, at equilibrium. [PH 3] eq = 0 1.33 x 10-4 23.1 3 0.0432 M 4 RESET 0.0432 8.27 x 10-3 3.06 x 10-3