Analysis of Statement 1

1. Statement 1: For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero.

2. Let's consider the relationship between the Gibbs energy of reaction ($\\\Delta_r G$) and the standard Gibbs energy of reaction (\\\Delta_r G^\\circ). The equation is given by: $\Delta_r G = \Delta_r G^\circ + RT \ln Q$ where $R$ is the gas constant, $T$ is the absolute temperature, and $Q$ is the reaction quotient.

3. At equilibrium, two conditions are met:

   • The Gibbs energy change for the reaction is zero: $\\\Delta_r G = 0$.
   • The reaction quotient $Q$ is equal to the equilibrium constant $K$: $Q = K$.

4. Substituting these equilibrium conditions into the equation, we get: $0 = \Delta_r G^\circ + RT \ln K$ Rearranging this gives the relationship between the standard Gibbs energy and the equilibrium constant: $\Delta_r G^\circ = -RT \ln K$

5. From this equation, we can see that the standard Gibbs energy of reaction, \\\Delta_r G^\\circ, is zero only if $\\\ln K = 0$, which implies that the equilibrium constant $K = 1$.

6. However, a chemical reaction can be at equilibrium with any value of $K$ (i.e., $K > 1$, $K < 1$, or $K = 1$). It is not necessary for $K$ to be 1 for every reaction at equilibrium.

7. Therefore, the statement that \\\Delta_r G^\\circ is zero for every chemical reaction at equilibrium is false. It is $\\\Delta_r G$ (the Gibbs energy change under the specific equilibrium conditions) that is zero, not necessarily \\\Delta_r G^\\circ (the standard Gibbs energy change).

Analysis of Statement 2

1. Statement 2: At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

2. This statement describes the fundamental criterion for spontaneity of a process under conditions of constant temperature and pressure. The change in Gibbs energy, $\\\Delta G$, determines the direction of a spontaneous process.

   • If $\\\Delta G < 0$ (negative), the process is spontaneous in the forward direction. A negative $\\\Delta G$ signifies a decrease in the system's Gibbs energy.
   • If $\\\Delta G > 0$ (positive), the process is non-spontaneous in the forward direction, but the reverse process is spontaneous.
   • If $\\\Delta G = 0$, the system is at equilibrium, and there is no net change.

3. Thus, a chemical reaction will proceed spontaneously in the direction that leads to a decrease in the Gibbs energy of the system.

4. Therefore, Statement 2 is true.

Conclusion

• Statement 1 is False.
• Statement 2 is True.

Based on this analysis, the correct option is the one that identifies Statement 1 as false and Statement 2 as true.