Now, notice we could have also used negative five joules or negative 1 million joules. Basically any negative value here works because any negative value means this reaction will be spontaneous under that Q situation, under that concentration of products and reactants. So if we take this equation now and solve for a log of Q, we get that log of Q is equal to so this term is brought to this side. So it's this minus this on the top. And then we divide by this quantity. So 2.33.