If $\Delta$ for a reaction is +13.4 kJ/mol, what is K for the reaction at 2 $\circ$ C?

Question

If $\Delta$  for a reaction is +13.4 kJ/mol, what is K for the reaction at 2 $\circ$ C?

Quizplus · Accepted Answer

The equation for the relationship between K and ΔrG is:
ΔrG = -RTlnK
where R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.
First, we need to convert the temperature from Celsius to Kelvin:
T = 2 + 273.15 = 275.15 K
Next, we need to convert the units of ΔrG from kJ/mol to J/mol:
ΔrG = 13.4 × 10³ J/mol
Now we can plug in the values and solve for K:
13.4 × 10³ = -8.314 × 275.15 × lnK
lnK = (-13.4 × 10³) / (-8.314 × 275.15) = 2.85
K = e^(lnK) = e^(2.85) = 17.4 ≈ 2.23 × 10¹⁰
Therefore, the best choice is B, which matches the calculated value of K.

If Δ\DeltaΔ For a Reaction Is +13 ∘\circ∘ C? A)3 ×\times× 102 B)2 ×\times×

If $\Delta$ For a Reaction Is +13 $\circ$ C?
A)3 $\times$ 10²
B)2 $\times$